Question

For planning purposes, a manufacturer wanted to know the average number of items sold daily. The manufacturer assumed that the number would be 12.8 and had budgeted for that amount. Realizing that this assumption may not be valid, the manufacturer chooses a random sample of 36 days and the results showed an average daily sale of 15.3 items with a standard deviation of 6.8.

Is there sufficient evidence at the alpha = 0.5 level of significance to believe that the actual number is different than the assumption and what do you conclude? What is the P-value of your test statistic?

Answer 1

H0: = 12.8 , Ha: 12.8

Test statistics

t = - / (S / sqrt(n) )

= 15.3 - 12.8 / ( 6.8 / sqrt(36) )

= 2.21

This is test statistics value.

df = n - 1 = 36 - 1 = 35

From T table,

With test statistics 2.21 and df = 35 ,

p-value = 0.0337

Since p-value < significance level , We have sufficient evidence to reject H0.

We conclude that, we have enough evidence to support the claim that the actual number is different

than thee assumption.