The mean amount of time needed to thaw a particular frozen dinner in the microwave prior to convection cooking is claimed to be 5.5 min, with standard deviation of 0.95 min.To test if the actual thaw time is longer than this, we sample 35 frozen dinners and record the amount of time to thaw each one.The average thaw time is 5.8 min.What is the p-value from this test?
A) 0.3761
B) 0.6239
C) 0.9693
D) 0.0307

Question

Quizplus · Accepted Answer

We need to conduct a one-sample t-test, where:

Null hypothesis: The true mean thaw time is 5.5 min.
Alternative hypothesis: The true mean thaw time is greater than 5.5 min.

The test statistic is given by:
t = (sample mean - hypothesized mean) / (standard deviation / sqrt(sample size))
Plugging in the values:
t = (5.8 - 5.5) / (0.95 / sqrt(35)) = 1.72

The degrees of freedom = sample size - 1 = 34
Using a t-table or calculator, the p-value is found to be approximately 0.0307.
Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the true mean thaw time is longer than 5.5 min.

The Mean Amount of Time Needed to Thaw a Particular