One-Sample Hypothesis A manufacturer wants to increase the shelf life of a line

Question

One-Sample Hypothesis

A manufacturer wants to increase the shelf life of a line of cake mixes. Past records indicate that the average shelf life of the mix is 216 days. After a revised mix has been developed, a sample of nine boxes of cake mix had a mean of 217.22 days with a standard deviation of 1.20 days.

Answer the following:

a. Assume you are trying to determine if the shelf life has increased. Determine if the appropriate hypothesis test should be an upper tail, lower tail, or 2-tail test.

b. Determine the value of the test statistic.

c. Determine the critical z or t value using a level of significance of 2.5%.

d. Determine the p-value.

e. State your conclusion: Reject Ho or Do not Reject Ho? (Using a significance level of 2.5%).

f. What is the probability of making a type I Error?

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu = 216 Vs H1 : mu > 216

where mu is populaiton mean shelf life of the mix.

It is one tailed upper tail test.

Assume alpha = level of significance = 2.5% = 0.025

Given that,

sample size (n) = 9

Sample mean (Xbar) = 217.22

standard deviation (s) = 1.20

Here sample size < 30 and population standard deviation is unknown so we use one sample t-test.

The test statistic is,

t = (Xbar - mu) / (s / sqrt(n))

t = (217.22 - 216) / (1.20 / sqrt(9)) = 3.05

Now we have to find P-value for taking decision.

P-value we can find in EXCEL.

syntax :

=TDIST(x, deg_freedom, tails)

where x is absolute value of test statistic.

deg_freedom = n-1 = 9-1 = 8

tails = 1

P-value = 0.0079

P-value < alpha (0.025)

Reject H0 at 0.025 level of significance.

Conclusion : There is sufficient evidence to say that shelf life has increased.

Type I error = P(Reject H0 / H1 : mu = 222)

= P(mu > 216 / mu = 222)

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