Assume that the population of weights of people (in kgs.) has an approximately n

Question

Assume that the population of weights of people (in kgs.) has an approximately normal

distribution with unknown mean µ and standard deviation = 10. 0. This question first

concerns the z-test for testing the null hypothesis H0 : µ = 60. 0 against the alternative

hypothesis H1 :µ < 60. 0 using = 0. 05, and second a confidence interval for µ.

Consider using a z-test along with = 0. 05 to test the alternatives H0 : µ= 60. 0,

H1 : µ < 60. 0.

Suppose that a random sample of n = 25 people was chosen, and their weights analyzed

by a statistical software package. Part of the output is as follows.

Test of H0 : µ = 60. 0 versus H1 : µ < 60. 0 .

N    mean      z-value    p-value

25   58. 0      1. 0        0. 1587

(i) [3 marks] Based on the printout above, should H0 be rejected? Give briefly a reason

for your answer.

(ii) Define the p-value of any statistical test, and for the output above,

show how the p-value can be calculated from the z-value.

(iii) Calculate the 95% confidence interval x +- 1. 96

for µ, and interpret it carefully.

(iv) Suppose that in the present problem, standard deviation was not known and that in

addition we were allowed to sample only 4 people (instead of 25). Then to carry out

the test above, i.e. to test if the population mean weight is less than 60, it would not be

appropriate to use the z-test. If the sample of 4 people showed a mean weight of 56 and

a standard deviation of 8. 0, calculate the value of the appropriate t-test statistic. What

is the critical point for the t-test if a level of significance = 0. 01 is used. What is the

decision?

Explanation / Answer

(i) Based on the printout above, should H0 be rejected? Give briefly a reason for your answer.

No, Null Hypothesis should not be rejected. As p - value is greater than alpha = 0.05 so it is not under significance level so we cannot reject the null hypothesis.

ii) Define the p-value of any statistical test, and for the output above, show how the p-value can be calculated from the z-value.

P - value here is probability of the mean of the sample is less than or equal to 58 when population mean = 60 and standard error of the sample mean is s/n .

so p - value = Pr( xbar <= 58; 60.0; 10.0/25) = P(xbar <= 58; 60.0; 2)

so Z - value = (58 - 60)/ 2 = -1

so p - value from Z -table = 0.1587

(iii) Calculate the 95% confidence interval x +- 1. 96 for µ, and interpret it carefully.

95% confidence interval for µ = x+- 1.96 * s/n = 58.0 +- 1.96 * (10/25) = 58 +- 1.96 *2 = (54.08, 61.92) and we can interprete that null hypothesis population mean is 60 is under 95% confidence interval for sample mean.

(iv) HEre sample mean xbar = 56

standard deviation of the sample s = 8

alpha = 0.01

here t test would be appropriate instead of Z test

so test statistic

t* = (xbar - µ)/ (s/n) = (56 - 60) / (8/4) = -4/ 4 = -1

so critical value of t for alpha = 0.01 and dF = 4-1 =3 is equal to 4.541

so here t* < tcritical so we cannot reject the null hypothesis and can conclude that population mean of weight of the people is equal to 60 kg,.

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