Question

In: Statistics and Probability

1.) A simple random sample of 150 M&M’s was collected with 30 blue M&M’s. M&M’s claims...

1.) A simple random sample of 150 M&M’s was collected with 30 blue M&M’s. M&M’s claims that it is greater than or equal to 24%. Use a 4-step procedure and conduct a hypothesis test using a .05 level of significance.

2.) Scientists gathered a simple random sample of the nicotine levels from 100 menthol cigarettes and the nicotine levels from 100 nonmenthol cigarettes. The simple random sample of 100 menthol cigarettes has a sample mean nicotine amount of 23.8 mg and a sample standard deviation of 32.0 mg. A simple random sample of 100 nonmenthol cigarettes has a sample mean nicotine amount of 47.6 mg and a sample standard deviation of 23.5 mg. Use a .05 significance level to test the claim that menthol cigarettes and nonmenthol cigarettes have different amounts of nicotine. Use our 4 step procedure.

Solutions

Expert Solution

1) Step 1

The null and alternative hypothesis

Step 2

Test criteria :

This is an one tailed test (left tailed)  as alternative hypothesis is p<0.24

For , one tailed critical value of z is , zc = - 1.65 ( from z table)

Reject H0 if z < -1.65

Step 3:

Test statsitic

where , 30/150 = 0.2

thus ,

Step 4 : Decision

Since z > -1.22

We fail to reject H0

At 0.05 level of significance , there is sufficient evidence to reject the claim that proportion of blue MMs greater than 0.24.

2) Step 1

The null and alternative hypothesis

Step 2

Test criteria

This is a two tailed t test

degrees of freedom = n1+n2-2 = 100=100-2=198

For , two tailed critical value of t is , tc = 1.97 (from t table)

Reject H0 if I t I  > 1.97

Step 3

Test statistic

where

23.8

47.6

32.0

23.5

= 28.0736

Thus

Step 4 : Decision

Since I t I > 1.97

We reject H0

At 0.05 level of significance ,there is sufficient evidence to conclude that support the claim that menthol cigerettes and non menthol cigerettes have different amount of nicotene.


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