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 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.
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.