Question

4. A smartphone carrier’s data at airports was collected. The smartphone carrier wants to test the claim the average speed is less than 5.00 Mbps. Use a .05 significance level. Assume the population is normally distributed. The sample mean is 4.42 and the sample standard deviation is 1.85. The sample size is 40.

Hypothesis:

Test Statistic and value:

p value:

Conclusion:

Answer 1

Solution :

Hypothesis :

The null and alternative hypotheses would be as follows :

Test statistic and value :

To test the hypothesis the most appropriate test is one sample t-test. The test statistic is given as follows :

Where, x̅ is sample mean, s is sample standard deviation, n is sample size and μ is hypothesized value of population mean under H_{​​​​​​0}.

We have,  x̅ = 4.42, s = 1.85, n = 40 and μ = 5.00

The value of the test statistic is -1.9828.

P-value :

Since, our test is left-tailed test, therefore we shall obtain left-tailed p-value for the test statistic. The left-tailed p-value is given as follows :

P-value = P(T < t)

P-value = P(T < -1.9828)

P-value = 0.0272

The p-value is 0.0272.

Conclusion :

P-value = 0.0272 and Significance level = 0.05

(0.0272 < 0.05)

Since, p-value is less than the significance level of 0.05, therefore we shall reject the null hypothesis (H_{​​​​​​0}) at 0.05 significance level.

At 0.05 significance level, there is sufficient evidence to support the the claim that the average speed is less than 5.00 Mbps.

Please rate the answer. Thank you.