Question

The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales representatives make an average of 41 sales calls per week on professors. Several reps say that this estimate is too low. To investigate, a random sample of 38 sales representatives reveals that the mean number of calls made last week was 42. The standard deviation of the sample is 3.9 calls. Using the 0.025 significance level, can we conclude that the mean number of calls per salesperson per week is more than 41? H0 : μ ≤ 41 H1 : μ > 41 1. Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic 2. What is your decision regarding H0? The mean number of calls is than 41 per week.

Answer 1

Answer:

Step-by-step explanation:

For the null hypothesis,

H0 : μ ≤ 41

For the alternative hypothesis,

H1 : μ > 41

Looking at the alternative hypothesis, this a a right tailed test.

Since no population standard deviation is given, the distribution is a student's t.

Since n = 38,

Degrees of freedom, df = n - 1 = 38 - 1 = 37

t = (x - µ)/(s/√n)

Where

x = sample mean = 42

µ = population mean = 41

s = samples standard deviation = 3.9

The test statistic would be

t = (42 - 41)/(3.9/√38) = 1.58

We would determine the p value using the t test calculator. It becomes

p = 0.28

Since alpha, 0.025 < than the p value, 0.28, then we would fail to reject the null hypothesis. Therefore, At a 2.5% level of significance, the sample data did nor show significant evidence that the mean number of calls is more than 41 per week.

Answer 2

Answer:

1. t - statistics =  1.581

2. We accept null hypothesis that  the mean number of calls is less than or equal to, i.e., ≤, 41 per week.

Step-by-step explanation:

Significance level = 0.025

Sample size = 38

Sample mean = 42

Standard deviation = 3.9

Claimed mean = 41

1. Compute the value of the test statistic. (Round your answer to 3 decimal places.)

$t-statistics = \frac{42 - 41}{3.9/\sqrt{38} }$

t - statistics =  1.581

2. What is your decision regarding null hypothesis? The mean number of calls is than 41 per week.

Degree of freedom = Sample size - 1 = 38 - 1 = 37

At 0.025 significance level,

t-table = 2.021

Since t - statistics (i.e. 1.581) is less than t-table (2.021), we accept null hypothesis that  the mean number of calls is less than or equal to, i.e., ≤, 41 per week.