Question

Suppose the values

1.39 2.22 2.38 1.60 1.50

are a random sample from a distribution assumed to be Normal but for which the mean and variance are unknown.

(a) Give a 95% confidence interval for the mean.

(b) Test the hypotheses H0: μ = 2.5 versus H1: μ ̸= 2.5 at the 0.05 significance level.

(c) Give the p-values for testing the following hypotheses: i. H0: μ=2.5versusH1: μ̸=2.5 ii. H0: μ≤2.5versusH1: μ>2.5 iii. H0: μ≥2.5versusH1: μ<2.5 Page 2 iv. State whether a 99% confidence interval for μ based on this sample would contain the value 2.5.

Answer 1

Values
1.39	0.183
2.22	0.162
2.38	0.316
1.6	0.048
1.5	0.101

Sample Mean   = 9.09 / 5 = 1.82

Sample Variance = .81 / (5-1) = = 0.20

Sample Standard Deviation = = 0.45

Part a)

95% Confidence Interval

= 2.571

(1.82 - 2.571 * (0.45 / ) < < (1.82 + 2.571 * (0.45 / )

( 1.303 < < 2.337 )

Part b) To test

H₀ :-

H₁ :-

Teest Statistic :-

t = (1.82 - 2.5) / (0.45 / )

t = -3.38

Test Criteria :-

Reject Null hypothesis (H₀ ) if, | t | > t_{/2, n-1}

t_{/2, n-1} = t _{0.025, 5-1} = 2.571

| -3.38 | > 2.571 i.e 3.38 > 2.571, hence reject null hypotheisis

Conclusion :- We Accept Alternative Hypothesis at 0.05 level of significance

Part C) i P-values for testing  H0: μ=2.5 versusH1: μ̸=2.5

Solution :- We already calculate the| t | statistic value = 3.38

Check the value 3.38 in t table at 4 degree of freedom ( number of beservation - 1 = 5-1 = 4)

In the table we get to vallues for 4 DF i.e 2.2776 - 3.747 for = ( 0.05 - 0.02)

Since it is two tail, values will appear both side and P valuw will become double

0.01 * 2 < < 0.025*2 = 0.02 < < 0.05

P value by technology = 0.029

ii). H0: μ≤2.5versusH1: μ>2.5

Here Alternative hypothesis if greater than (right side of the graph)

So P value lies between 0.01 < P value < 0.025

P value by technology = 0.0140

iii. H0: μ≥2.5versusH1: μ<2.5

Here Alternative hypothesis is left sided and in the t table we always get right side area of the region

hence we need to subtract 1 from the area of the right side P value

Right side P value = 0.01 < P vlaue < 0.025

Left Side P value = 1 - 0.01 < P value < 1 - 0.025 = 0.99 < P value < 0.975

P value by technology = 0.986

iv) State whether a 99% confidence interval for μ based on this sample would contain the value 2.5.

99% Confidence Interval

= 4.604

(1.82 - 4.604* (0.45 / ) < < (1.82 + 4.604* (0.45 / )

( 0.893 < < 2.747 )

Value 2.5 lies in the interval.