1. ## Statistics Questions

A grocery store produce manager is told by a wholesaler that the apples in a large shipment have a mean weight of 7 ounces and a standard deviation of 1.4 opunces. The mager is gloing to randomly select 49 apples and is willing to risk a 2.5% chance of returning the shipment, assuming the wholsaler’s claim is true and the sample mean weight of the 49 apples falls below x ounces. Find x.

An executive member of an ice cream company claimed that its product contained more than 500 calories per pint. To test this claim, 81 containers were randomly chosen and analyzed. The sample result showed that the mean was 509 calories with a standrad deviation of 27 calories. (a.) At α=5%, using the classical approach to test the claim. (b.) Using the P-value approach to conclude it.

A pyschological test inventory called POMS was given to a sample of 25 runners. The result showed that the mean was 10.46 with standard deviation of 5.55. A psychologust suspected that, in general, runners have a mean score on POMS that is different from there national norm of 13. (a.) At α=1%, using the classical approach to test the claim. (b.) Using the P-value appraoch to conclude it.

The Speedy oil change company advertised the mean completion time for an oil change is less than 15 minutes. A sample of 16 oil chnages resulted in a mean of 14 minutes and a standrad deviation of 4 minutes. (a.) At α=10%, using the classical appraoch to test the claim. (b.) Using the P-value appraoch to conclude it.

2. Originally Posted by cantdomath
The Speedy oil change company advertised the mean completion time for an oil change is less than 15 minutes. A sample of 16 oil chnages resulted in a mean of 14 minutes and a standrad deviation of 4 minutes. (a.) At α=10%, using the classical appraoch to test the claim. (b.) Using the P-value appraoch to conclude it.
n=16
X = 14
sigma = 4
null: mean >= 15
alternative: mean < 15

For part a)

Find the z-score (one-tail) for a p-value of 0.10. On the ti-83, invNorm(0.10) gives -1.2815.

Find the z-score $= \frac{X - mean}{\frac{sigma}{\sqrt{n}}} = -1$

Compare the z-score to -1. If it's less than -1, that means the observation of 14 minutes is extreme, so you reject the null and accept the alternative.

For part b) Find the P-value of the z-score of -1 and compare it to the significance level.
I hope this helps.