Question

A head librarian supervises a number of libraries in a large county. He wants to know if full-time library workers and part-time library workers re-shelve books at the same rate. So, he checks the records of 45 full-time library workers and finds that they re-shelve an average of 166 books per hour with a standard deviation of 9.3 books per hour. The records of 45 part-time library show that they re-shelve an average of 159 books per hour with a standard deviation of 12.2 books per hour.

Using a level of significance of α=.01, is there enough evidence to indicate a difference in the mean number of books re-shelved by full-time workers compared to part-time workers?

please be detailed in telling how to locate the z= 1.645 and z= -1.645

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 2.2868
DF = 88
t = [ (x₁ - x₂) - d ] / SE

t = 3.06

t_Critical = + 2.633

Rejection region is - 2.633 > t > 2.633

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Interpret results. Since the t-value (3.06) lies in the rejection region, hence we have to reject null hypothesis.

From the above test there is enough evidence to indicate a difference in the mean number of books re-shelved by full-time workers compared to part-time workers.