Question

1. Last year, the mean amount spent by customers at a local sandwich shop was $18.50. The shop owner believes that the mean may be higher this year. State the appropriate null and alternate hypotheses.

2. A test is made of H0 : m = 15 versus H1 : m < 15. The true value of m is 13 and H0 is not rejected. Is this a Type I error, a Type II error, or a correct decision?

3. A test is made of H0:m=20versus H1:m>20. The value of the test statistic is z=1.87. Do you reject H0 at the a=0.05level? Do you reject H0 at thea=0.01 level?
a = 0.05 _________

a = 0.01 _________

4. A certain type of calculator battery has a mean lifetime of 120 hours and a standard deviation of s = 12 hours. A company gas developed a new battery and claims it has a longer mean life. A random sample of 500 batteries is tested, and the sample mean lifetime is x = 121 hours. Can you conclude that this new battery has a longer mean life? Use the a= 0.01 level. Provide the null and alternate hypotheses, compute the value of the test statistic, and state a conclusion.

5. In 2012, the mean number of runs scored by both teams in a Minor League Baseball game was 7.72. In 2013, a sample of 32 games yielded a mean number of runs of 7.09 with a standard deviation of 2.85. Perform a hypothesis test to determine whether the mean number of runs in 2013 is less than it was in 2012. Use the a = 0.05 level. Provide the null and alternate hypotheses, compute the value of the test statistic, and state a conclusion.

z = x - m0 sn

Table of critical values:

Hypothesis testing: test statistic for a mean, standard deviation unknown:

t= x-m0 Use Table C with df=n-1for critical values

	a = 0.05	a = 0.01	a = 0.10
left-tailed	-1.645	-2.326	-1.282
right-tailed	1.645	2.326	1.282
2-tailed	±1.96	±2.576	±1.645

Answer 1

1. We are testing,

H0: u= 18.5 vs H1: u>18.5

2. Here m=13, so the null hypothesis should have been rejected. But since H0 is not Rejected we have committed a Type II error here(non rejection of a false null hypothesis)

3. Critical value of this one sided z test is at z0.05= 1.645

Since the test statistic of 1.87 > critical value here, we have sufficient evidence to Reject H0 at the 5% levsl of significance.

At alpha=0.01, critical value is z0.01= 2.33

Here since test statistic of 1.87 <critical value, we fail to Reject H0 at the 1% level of significance.

4. We are testing,

H0: u=120 vs H1: u>120

Under H0, test statistic: (121-120)/(12/√500) = 1.863

Critical value of this one sided z test for alpha =0.01 is at z0.01= 2.33

Since the test statistic of 1.863<test statistic, we have insufficient evidence to Reject H0 at the 1% level of significance. So we fail to Reject H0. Hence we fail to conclude that the mean lifetime has improved.