STAT-C 316 (NONPARAMETRIC STATISTICS)

Chapter 02 - One Sample Nonparametric Methods

INTRODUCTION

In classical parametric test (which assume that the population from which the sample data have been drawn is normally distributed), the parametric of interest is the population mean. In this chapter, we shall be concerned with the nonparametric analog of the one-sample z and t tests. These are nonparametric procedures (which utilize data consisting of a single set of observations) that are appropriate when the location parameter is the median, rather than the mean.

Several nonparametric procedures are available for making inferences about the median. Two of the nonparametric tests which are useful in situations where the conditions for parametric z and t tests are not me, are the one-sample sign test and the Wilcoxon Signed-ranks test.

Recall that the median of a set of data is defined as the middle value when data are arranged in order of magnitude. For continuous distributions, we define the median as the point μ~ for which the probability that a value selected at random from the distribution is less than μ~, and the probability that the value selected at random from the distribution is greater than the μ~ are both equal to 1/2.

When the population from which the sample has been drawn is symmetric, any conclusions about the median are applicable to the mean, since in symmetrical distributions the mean and the median coincide.

In this chapter, we shall also discuss procedures for making inferences concerning the population proportion and testing for randomness and the presence of trend.

Wherever possible, we shall observe the following format in presenting the hypothesis-testing procedures.

Assumptions.

We list the assumptions necessary for the validity of the test, and describe the data on which the calculations are based.

Parameter of interest.

From the problem context, we identify the parameter of interest

Hypotheses.

We state the null hypothesis H₀ and the alternative hypothesis H₁

Test Statistic.

We write down a formula or direction for computing the relevant test statistic. When we give a formula, we describe the methodology for evaluating it.

Significance level.

We choose a significance level .

Decision rule

We determine the critical region. The Appendix gives appropriate tables for the distribution of the test statistic. From these tables, we can determine the critical values of the test statistic corresponding to the chosen .

Value of the test statistic

We compute the value of the test statistic from the sample data.

Decision

If the computed value of the test statistic is as extreme as or more extreme than a critical value, we reject H₀ and conclude that H₁ is true.

If we cannot reject H₀, we conclude that there is not enough information to warrant its falsity.

The One-Sample Sign Test

The sign test is perhaps the oldest of all nonparametric procedures. Let X₁, X₂,...Xn be an observed random samples of size n from a population with median ῠ. The sign test utilizes only the signs of the differences between observed values Xi and the hypothesized median ῠ₀, Thus the data is converted into a series of plus (+) and minus (-) signs.

Assumptions:

The sample available for analysis is a random sample of independent measurements from a population with an unknown median ῠ.
The variable of interest is measured on at least an ordinal scale
The variable of interest is continuous

Hypothesis:

The hypothesis to be tested concerns the values of the population median. To test the hypothesis Ho: ῠ = ῠ₀

where ῠ₀ is a specified median value, against the corresponding one-sided or two-sided alternative, we use the sign test. The test statistics S depends on the alternative hypothesis, H₁

(a) One-sided Test:

For a one-sided test, the alternative hypothesis is either H₁: ῠ < ῠ₀ or H₁: ῠ > ῠ₀

(i) if we wish to test,

Ho: ῠ = ῠ₀ against, H₁: ῠ < ῠ₀ then the t statistics is defined by S = N+,

where N+ = Number of observations Xi greater than ῠ₀

= Number of +signs when the differences Xi - ῠ₀ are computed, i = 1, 2, ..., n.

if the alternative hypothesis is true, then we should expect Xi - ῠ₀ to yield significantly fewer positive (+) signs than negative (-) signs. Thus, a smaller number of (+) signs leads to the rejects of H₀, when H₀ is true, we expect the number of (-) signs to be equal to that of the (+) signs and hence P(S<ῠ₀) = P(S>ῠ₀) = ½.

Thus, when H₀ is true, S has the binomial distribution with parameters n and ½ that is S ~ b(n, ½) [read as S is a binomial random variable with parameters n and 1/2]

Decision Rule:

The P-value of the test is defined by

p = P(S ≤ s₀ | H₀ is true),

Where s₀ is the observed value of the test Statistics S. We reject H₀ at significance level α if p ≤ α

(ii) For a one-sided test, we test

H₀: ῠ = ῠ₀ against

H₁: ῠ > ῠ₀

The test Statistics is S = N -

Where N- = Number of observations less than ῠ₀

= Number of (-) signs when the differences Xi - ῠ₀ are computed, i = 1, 2, ... n.

If the alternative hypothesis is true, then we should expect Xi - ῠ₀ to yield less negative (-) signs than would be expected if the null hypothesis were true. Likewise, when H₀ is true, S has the binomial distribution with parameters n and ½. That is S ~ b(n, ½).

Decision Rule

The p-value of the test is defined by

p = P(S ≤ s₀ | H₀ is true),

Where s₀ is the observed value of S = N-. We reject H₀ at significance level α if p ≤ α.

(b). Two-sided test.

If we wish to test,

H₀ : ῠ = ῠ₀ against,

H₁: ῠ ≠ ῠ₀ then the t statistics is defined by

S = min { N -, N+),

Where N- is the number of (-) signs and N+ is the number of (+) signs when the differences Xi - ῠ₀ are computed.

We should reject the null hypothesis if we have too few negative (-) signs or too few positive (+) signs. When H₀ is true, S has the binomial distribution with parameters n and ½.

Decision Rule

The p-value of the test is defined by

p = 2P(S ≤ s₀ | H₀ is true),

Where s₀ is the observed value of the test Statistics. We reject H₀ at significance level α if p ≤ α.

Problem with zero (0) differences

We assume that the variable of interest is continuous. Therefore, in theory no zero differences should occur when we compute Xi - ῠ₀.
In practice, however, zero differences do occur. The usual procedure is to discard observations leading to zero differences and reduce n accordingly. In that case the hypothesis may be re-stated in probability terms. For example, a two-sided case will have its null hypothesis as P(X < ῠ₀.) = P(X>ῠ₀.) = 0.50

Example 1.

Appearance transit times for 11 patients with significantly occluded right coronary arteries are given below:

Can we conclude, at the 0.05 level of significance, that the median appearance transit time in the population from which the data were drawn, is different from 3.50 seconds?

Solution

The parameter of interest is ῠ, the median appearance transit time in the population. We wish to test hypothesis

H₀ : ῠ = 3.50 against

H₁ : ῠ ≠ 3.50,

at the α = 0.05 level of significance. Since this a two-sided test, the test statistics is S = min { N-, N+}

Where N- is the number of observations less than 3.50, and N+ is the number of observations greater than 3.50. When Ho is true, S ~ b(10, 0.5)

Note: We discard one observations which has the same value as the hypothesized median, leaving us with a usable sample size of 10.

Let s₀ be the observed value of the test statistics, We reject H₀ at the 0.05 level of significance when p ≤ 0.05, where p = 2P(S ≤ s₀ |10, 0.5)

From the above table, N- = 9 and N+ = 1. The observed value of test statistics is therefore given by s₀ = min {9, 1} = 1

Since this is a two sided test, the p-value of the test is given by p= 2P (S ≤ 1 | 10, 0.5) = 2 x 0.0107 = 0.0214.

since the p-value of the test, 0.0214, is less than 0.05, we reject Ho at the 0.05 level of significance and conclude that the population median is not equal to 3.50.

Extra:

BINOMIAL DISTRIBUTION FORMULA

Example 2.

Solution

Large Sample Approximation

if the sample is larger than 15, we can use the normal approximation to the binomial distribution with a continuity correction. Thus, if n is large and S ~ b(n, ½), then it can shown that S is approximately normally distributed with mean np and variance np (1-p).

That is, S ~ N(np, np(1-p)). Thus, for the sign test, when p = ½ and n > 15, we can use the test statistics

When H₀ is true and n ≥ 15, Z is approximately N(0,1). For the large sample approximation, it is common to use continuity correction, by replacing S by S+½ in the definition of Z. Equation 1 above becomes

Example 3

The following data gives the ages, in years, of a random sample of 20 students from Besease Senior High School. It is believed that the median age of students in this school is smaller than 22 years. Based on these data, is there a sufficient evidence to conclude that the median age of students from Besease Senior High School is smaller than 22 years?

Solution:

The parameter of interest is ῠ, the median age of students from Besease Senior High School. We are interested in testing the null hypothesis

H₀ : ῠ = 22 against

H₁ : ῠ < 22,

The test statistics is S = N+

Where N+ = number of observations Xi greater then 22

= number of signs when the differences Xi - 22 are computed, i = 1, 2,...,20.

When H₀ is true, Z is N(0,1). Let z₀ denote the observed value of the test statistics Z. We reject Ho at the 0.05 level of significance when z₀ ≤ zα = z0.05 = -1.645. The following table gives the signs of Xi - 22.

From the above table, N+ = 5. Thus, the observed value of the statistics S is 5. This gives,

Since -2.0125 is less than -1.645, we reject H₀ at the 0.05 level of significance and conclude that the median age of students of Besease Senior High School is less than 22 years.

Confidence interval for the median based on the sign test

The 100(1-α) confidence interval for ῠ consists of those values of ῠ₀ for which we would not reject a two-sided null hypothesis H₀: ῠ = ῠ₀ at the α level of significance.
We designate the lower limit of our confidence interval by ῠL and the upper limit by ῠu.
We determine the largest positive or negative signs (i.e. the values of s') such that P (S ≤ s' | n, 0.05 ) = α/2
When the data values are arranged in order of magnitude, the (s' + 1)th observation is ῠL. To find ῠu, the upper limit of the confidence interval, we count the ordered sample values backward from the largest. The (s' + 1)th observation from the largest value locates ῠu. i.e. ῠu = (n - s')th value.

Example 4.

Construct a 95% confidence interval for the median of the population from which the following sample data have been drawn, using the sign test.

Solution:

The point estimate of the population median is the sample median which is the mean of two middle values is the ordered array, thus.

The sample median = (3.08+3.10)/2 = 3.09

To find ῠL, we consult a table of the binomial distribution and find that P(S ≤ 3 | 16, 0.5) = 0.0105 and P(S≤ 4 | 16, 0.5) = 0.0383.
Thus, we note that we cannot obtain exact 95% confidence interval for the median. Since 100[1-2(0.0105)] = 97.9, which is larger than 95 and 100[1-2(0.0383)] = 92.34, which is smaller than 95.
This method of constructing confidence intervals for the median does not usually yield intervals with exactly the usual coefficients of 0.90, 0.95, and 0.99.
In practice, we choose between a wider interval and a higher confidence of the narrower interval and lower confidence.
Suppose we choose s' = 4, then s' + 1 = 5. Therefore the 5th value in the ordered array is ῠL and the 12th (i.e. 16 - 4) value in the ordered array is ῠu.
Thus ῠL = 1.99 and ῠu = 4.01.
The confidence coefficient is therefore 100[1-2(0.0383)] = 92.34. We say that we are 92.34% confident that the populations median is between 1.99 and 4.01.

Large Sample Approximation

Example 5.

Construct a 95% confidence interval for ῠ. Refer to the data below.

Solution

Exercise.

Solution

2. The Wilcoxon Signed-ranks test

As we have seen, the sign test utilizers only the signs of the differences between observed values and the hypothesized median. For Testing H₀: ῠ = ῠ₀, there is another procedure that uses the magnitude of the differences when these are available.

The Wilcoxon signed-ranks procedures makes use of additional information to rank the difference between the sample measurements and the hypothesized median.

The Wilcoxon signed-ranks uses more information than the sign test, making it more powerful test when the sampled population is symmetric. However, the sign test is preferred when the sampled population is not symmetric.

Assumptions

The sample available for analysis is a random sample of size n from a population with an unknown median ῠ
The variable of interest is measured on a continuous scale
The sampled population is symmetric
The scale of measurement is at least interval
The observations are independent

Hypothesis

The parameter of interest is ῠ, the population median. To test the hypothesis H₀: ῠ = ῠ₀

Where ῠ₀ is the hypothesized median, against a corresponding one-sided or two-sided alternative, we can also use the Wilcoxon signed-ranks test.

Test Statistics

To obtain the test statistic, we use the following procedure.

Subtract the hypothesized median, ῠ₀ from each observation Xi, that is, for each observation Xi, find:

Di = Xi - ῠ₀, ∀ i = 1, 2,..., n

If any observation Xi is equal to the hypothesized median, ῠ₀, eliminate it from the calculations and reduce the sample size accordingly.
Rank the differences | Di |, from the smallest to largest without regard to their signs. if two or more |D1| are tied, assign each tied value the mean of the rank positions of the tied differences.
Assign to each rank the sign of the differences of which it is ranked.
Obtain the sum of the ranks with positive signs; call it W+. Obtain the sum of the ranks with negativr signs; call it W-.
Note that:

For a given sample, we do not expect W+ to be equal to W-

Carrying out the Wilcoxon signed ranks test.

When the null hypothesis,

Ho : ῠ = ῠ₀

is true, we do not expect great difference between W+ and W-. Consequently, a sufficiently small value of W+ or a sufficiently small value of W- causes us to reject Ho.

(a) One-sided test: To test,

Ho : ῠ = ῠ₀, against

H₁ : ῠ < ῠ₀

at the α level of significance.

Test Statistics

A sufficiently small value of W+ leads to the rejection of the null hypothesis Ho. The test statistics therefore is W = W+.

Decision Rule

We reject Ho at significance level α if the observed W value w₀, is less than or equal to the tabulated W value n and a preselected α.

(b) One-sided test: To test,

Ho : ῠ = ῠ₀, against

H₁ : ῠ > ῠ₀

at the α level of significance.

Test Statistics

For a sufficiently small W- value, we reject Ho. The test Statistics therefore is W = W-,

Since a small value causes us to reject the null hypothesis.

Decision Rule

We Reject Ho at significance level α if the observed W value, w₀, is less than or equal to the tabulated W value for n and a preselected value of α.

(c) For a two-sided test, we test

Ho : ῠ = ῠ₀, against

H₁ : ῠ ≠ ῠ₀

at the α level of significance.

Test Statistics

The test statistics is

W = min(W-, W+)

Since a small value of either W- or W+ causes us to reject the null hypothesis.

Decision Rule

We reject Ho at significance level α if the observed W value, w₀, is less than or equal to the tabulated W value for n and a preselected value of α/2.

The distribution of W

The smallest value W can take is zero (0) and the largest value W can take is the sum of the integers from 1 to n: that is (n(n+1))/2. W is therefore a discrete random variable whose support ranges between 0 and (n(n+1))/2.
It can be shown that the probability mass function of the discrete variable W is given by

where c(w) - the number of possible ways to assign +sign or a -sign to the first n integers so that the sum of the ranks with +signs (or -signs) is equal to w.

Example 6.

The following are the systolic blood pressures (mmHg) of 13 patients undergoing a drug therapy for hypertension:

Can we conclude on the basis of these data that the median systolic blood pressure is less than 165 mmHg? Take α = 0.05

Solution:

The parameter of interest is median, ῠ, The median systolic blood pressure of the population. We wish to test the hypothesis

Ho : ῠ = 165, against

H₁ : ῠ < 165

at the α = 0.05 level of significance. Using the Wilcoxon Signed Ranked Test, the test statistics is W = W+

Where W+ is the sum of the ranks with positive signs.

We reject Ho at the 0.05 level of significance if w ≤ w₁₂,₀.₀₅ = 13, where w₀ is the observed value of the test statistics.

From Table A, W- = 50.5 and w+ = 27.5. The value of the test statistics is therefore w₀ = 28.5. Since 28.5 > 13, we fail to reject Ho. We conclude that the median systolic blood pressure of the subjects in the population is not less than 165 mmHg.

Wilcoxon Signed-ranks Distribution Table

Example 7.

Refer to Example 2. Use the Wilcoxon signed-ranks test to determine if there is any evidence that the median IQ of drug abusers in the population is different from 107. Use α = 0.05

Solution

Let ῠ denote the median IQ of drug abusers who are aged 16 years or older. We wish to test the hypothesis.

Ho : ῠ = 107, against

H₁ : ῠ ≠ 107.

at the α = 0.05 level of significance. The test statistics is W = min (W-, W+)

Where W- and W+ are the sums of the ranks with negative and positive signs, respectively.

We reject Ho at the 0.05 level of significance if w₀ ≤ w₁₄,₀.₀₂₅ = 17, where w₀ is the observed value of the test statistics.

From Table B. W- = 40.5 and W+ = 64.5. The value of the test statistic is w₀ = 40.5.

Since 40.5 > 17, we fail to reject Ho. We conclude that the median IQ of the subjects in the population may be 107.

Large Sample Approximation

Theorem 1:

Theorem 2.

Adjustment for ties

We can incorporate an adjustment for ties among nonzero differences in the large sample approximation in the following way.
Let t be the number of absolute differences tied for a particular nonzero rank. Then the correction factor is.

We can subtract this quantity from the expression in the denominator under the square root sign
Thus the adjusted statistics for a large sample approximation is

We illustrate the calculation of an adjustment for ties in the following data:

Example 8.

The following data show the life span, in years, of a random sample of 21 recorded deaths in a certain country. It has been known in the past years that the median life span in the country is 50 years. Can we conclude from these data that the median life span in the country has improved? Use α = 0.05.

PROBLEM SETS

Problem Set No. 01

The median age of the onset of diabetes is thought to be 45 years. The ages at onset of a random sample of 16 people with diabetes are given below:

26.2 30.50 35.50 38.00 39.80 40.30 45.00 45.60

45.90 46.80 48.90 51.40 52.40 55.60 60.90 65.40

Perform:

(a) sign test, (b) Wilcoxon signed-ranks test,

to determine if there is any evidence to conclude that the median age of the onset of diabetes differs significantly from 45 years. Use α = 0.05.

Problem Set No. 02

The following are the blood glucose levels of 12 patients who attend St. Thomas Hospital:

86 100 120 90 101 98 109 108 93 107 99 110

Perform the Wilcoxon signed-ranks test to determine if we can conclude on the basis of these data that the average glucose level in the population is greater than 96 mg/dl? Take α = 0.05.

3. The Binomial Test

Inferences concerning proportions are required in many areas. The population proportion is a parameter of frequent interest in research and decision-making activities.

The politician is interested in knowing what proportion of voters will vote for him in the next election.
All manufacturing firms are concerned about the proportion of defective items when a shipment is made.
A market analyst may wish to know the proportion of families in a certain area who have central air conditioning.
A sociologist may want to know the proportion of heads of household in a certain area who are women.

Many questions of interest to the health worker relate to the population proportion.

What proportion of patients who receive a particular treatment recover?
What proportion of a population has a certain disease?

When it is impossible or impractical to survey the total population, researchers base decision regarding population proportions, on inferences made by analyzing samples drawn from the population.

As usual, inference may take the form of interval estimation or hypothesis testing. Sometimes, we want to draw inferences concerning the total number, the proportion or percentage of units in the population that possess some characteristic or attribute or fall into some defined class.

A random sample of size n is drawn from a population. Suppose we wish to estimate the proportion, p, of units in the population that belong to some definite class in the population. Testing hypotheses about population proportions is carried out in much the same way as for median when the assumptions necessary for the test are satisfied.

Assumptions

Hypothesis

Example:

In a survey of injection drug users in a large city, Coates et al. (1991) found that 2 out of 12 were HIV positive. We wish to know if we can conclude, at the 10% level of significance, that fewer than 40% of the injection drug users in the sampled population are HIV positive.

Solution:

The parameter of interest is p, the proportion of injection drug users in the sampled population who are HIV positive.

We wish to test:

H₀ : p = 0.4, against

H₁ : p < 0.4

at significance level, α = 0.1. The test Statistics is S, the number of injection drug users in the sample who are HIV positive. When H₀ is true, S has the binomial distribution with parameters n = 12 and p = 0.4. Thus, S ~ b (12,0.4).

Let s₀ denote the observed value of the test statistics. We reject H₀ at the 0.1 level of significance if the p-value ≤ 0.1, where the p-value = P (S ≤ s₀ | 12, 0.4). Given s₀ = 2, p-value = P(S ≤ 2 | 12, 0.4) = 0.0834.

Since the p-value, 0.0834 < 0.1, we reject H₀ at the 10% level of significance and conclude that fewer than 40% of the injection drug users in the sampled population are HIV positive.

Example:

A researcher found anterior sub-capsular vacuoles in the eyes of 6 out of 15 diabetic patients.

Using binomial test, can we conclude that the population proportion with the condition of interest is greater than 0.20? Use α = 0.05.

Solution:

The parameter of interest is p, the proportion of diabetic patients in the population with anterior sub-capsular vacuoles in the eyes. We wish to test:

H₀: p = 0.2 against H₁: p > 0.2

The test statistics is S, the number of diabetic patients in the sample with anterior sub-capsular vacuoles in the eyes. When Ho is true,

S ~b(15, 0.2)

Let s₀ denote the observed value of the test statistics. We reject Ho at the 0.05 level of significance if the p-value of the test ≤ 0.05, where p-value = P ( S > s₀ | 15, 0.2). Given s₀ = 6,

p-value = P ( S > 6 | 15, 0.2) = 1 - P ( S > 6 | 15, 0.2) = 1 - 0.9819 = 0.0181.

So the p-value 0.0181 < 0.05, we reject H0. at the 0.05 level of significance and conclude that the population proportion p is greater than 0.2

Large Sample Approximation

If S is a binomial random variable with parameters n and p₀, then the expectation and variance of S are given by:

E(S) = np₀ and V(W) = np₀ (1- p₀).

Thus, when the null hypothesis is true, and n is large.

3. The normal approximation to the binomial distribution is good if np₀ > 5 and n(1- p₀) > 5.

4. Note that the sign test discussed earlier is a special case of the binomial test, in which p₀ = 0.5.

Example:

A commonly prescribed drug for relieving nervous tension is believed to be only 60% effective. Experimental results with a new drug administered to a random sample of 100 adults who were suffering from nervous tension show that 70 received relief. Is this sufficient evidence to conclude that the new drug is superior to the one commonly prescribed? Use α = 0.05.

Solution:

The parameter of interest is p, the proportion of adults in the population who received relief from nervous tension. We wish to test:

H₀: p = 0.6 against H₁: p > 0.6

at α = 0.05 level of significance. The test statistics is

Given n = 100 and p₀ = 0.6, both np₀ and n(1-p₀) are greater than 5 and so Z is approximately N(0,1) when Ho is true. We reject Ho if z, the computed Z value is greater than Z0.95 = 1.645. Now S = 70 and

Since 2.0412 > 1.645, we reject Ho at the 0.05 level of significance. We conclude that the new drug is superior to the one commonly prescribed.

Large Sample Confidence interval for p.

if p is the proportion of observations in a random sample of size n that belongs to a class of interest, then an approximate 100(1-α)% confidence interval of the proportion of the population that belongs to this this class is.

Where p = s/n is the proportion of the sample with the characteristics of interest.

Answer the following exercises below:

3. The one sample runs test for randomness

In many situations we want to know whether we can conclude that the set of observations constitute a random sample from an infinite population. Test for randomness is of major importance because the assumption of randomness underlies statistical inference. In addition, test for randomness are important for time series analysis. The runs test procedure is used to examine whether or not a sequence of sample values is random.

Consider, for example, the following sequence of sample values is random.

21 23 24 27 30 28 27 26 25 23 22 21

Each observations is denoted by a "+" if it is more than the previous observation and by a "-" sign if it is less than the previous observations as shown in the following table:

A run is a sequence of signs of the same kind bounded by signs of other kind. In this case, we doubt the sequence randomness, since there are only two runs.

if the order of occurrence were:

We would doubt the sequence's randomness because they are too many runs (10 in this instance)

Too few runs indicate that sequence is not random (has persistency) whilst too may runs also indicates that the sequence is not random.

Let us now consider the one sample runs test. This procedure helps us to decide whether a sequence of sample is the result of a random process.

Assumptions:

The data available for analysis consist of a sequence of sample values, recorded in the order of their occurrence.

Hypothesis:

We wish to test:

Ho: The sequence of sample values is random, against

H₁: The sequence of sample values is not random

Test Statistics:

The test statistics is R, the total number of runs.

Decision Rule:

Since the null hypothesis does not specify the direction, a two-sided test is appropriate. The critical value , rc, for the test is obtained from a table of critical values for a given sample size n, and at a desired level of significance α.

If rc (lower) ≤ r ≤ rc (upper), accept Ho. Otherwise reject Ho.

Tied Values:

If an observation is equal to its preceeding observations, denote it by zero. While counting the number of runs, ignore it and reduce the value of n accordingly.

Large Sample Size:

if n > 25, then the test statistics can be approximated by:

Example:

The following are the blood glucose levels of 12 patients who attend St. Thomas Hospital: Test, at 0.05 level of significance whether the sequence is random?

86 99 98 90 109 101 100 110 110 93 108 120

Solution:

Critical Values of r in the runs test

Exercise:

The following data show the average daily temperatures recorded at Accra, Ghana, for 15 consecutive days during June 2017.

Test, at the 0.05 level of significance, if we can conclude that the pattern of temperature is random?

2. The following data show the inflation rate in Ghana form 2006 to 2017. Test at the 0.05 level of significance, if we can conclude that the pattern of year inflation is random?

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