Question
1. H0:=210F0:=210F H1:210F1:210F Level of significance: =0.02=0.02 Sample size: n=11=11 Test statistic: z=0.48=0.48 Known population standard deviation Distribution of population data appears normal Standard Normal
1. H0:=210F0:=210F H1:210F1:210F Level of significance: =0.02=0.02 Sample size: n=11=11 Test statistic: z=0.48=0.48 Known population standard deviation Distribution of population data appears normal
Standard Normal Distribution Table
a. Determine the critical value(s) for the proposed hypothesis test.
- +
Round to two decimal places if necessary
b. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject
c. State the decision error type that is possible.
Type I Error
Type II Error
2. Test the claim that the average weight of a new SUV is 1,940 kg, if a sample of 287 vehicle weights results in a sample mean of 1,931 kg, with a standard deviation of 61.1 kg. Use a 2% level of significance.
Standard Normal Distribution Table
a. Calculate the test statistic.
z==
Round to two decimal places if necessary
b. Determine the critical value(s) for the hypothesis test.
- +
Round to two decimal places if necessary
c. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject
3. H0:=9,4640:=9,464
H1:<9,4641:<9,464
xsn=9,119=1,236=62=0.025x=9,119=1,236=62=0.025
T-Distribution Table
a. Calculate the test statistic.
t==
Round to three decimal places if necessary
b. Determine the critical value(s) for the hypothesis test.
- +
Round to three decimal places if necessary
c. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject
4. An office administrator for a physician is piloting a new "no-show" fee to attempt to deter some of the numerous patients each month that do not show up for their scheduled appointments. However, the administrator wants the majority of patients to feel that the fee is both reasonable and fair. She administers a survey to 34 randomly selected patients about the new fee, out of which 24 respond saying they believe the new fee is both reasonable and fair. Test the claim that more than 50% of the patients feel the fee is reasonable and fair at a 2.5% level of significance.
Standard Normal Distribution Table
a. Calculate the test statistic.
z==
Round to two decimal places if necessary
Enter 0 if normal approximation to the binomial cannot be used
b. Determine the critical value(s) for the hypothesis test.
- +
Round to two decimal places if necessaryEnter 0 if normal approximation to the binomial cannot be used
c. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject
Cannot Use Normal Approximation to Binomial
5. Conduct the stated hypothesis test for 12.12. Assume that the samples are independent and randomly selected from normal populations.
H0:12=00:12=0 | H1:1201:120 | =0.2=0.2 |
n1=361=36 | x1=2,242x1=2,242 | 1=154.51=154.5 |
n2=312=31 | x2=2,292x2=2,292 | 2=174.32=174.3 |
Standard Normal Distribution Table
a. Calculate the test statistic.
z==
Round to two decimal places if necessary
b. Determine the critical value(s) for the hypothesis test.
- +
Round to two decimal places if necessary
c. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject
6. Assume that the samples are independent and randomly selected from normal populations with equal population variances (12=22)(12=22).
T-Distribution Table
a. Calculate the test statistic.
t==
Round to three decimal places if necessary
b. Determine the critical value(s) for the hypothesis test.
- +
Round to three decimal places if necessary
c. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject
7. Determine if the conditions required for the normal approximation to the binomial are met. If so, calculate the test statistic, determine the critical value(s), and use that to decide whether there is sufficient evidence to reject the null hypothesis or not at the given level of significance.
H0:p=0.850:=0.85
H1:p0.851:0.85
pn=0.792=117=0.05p=0.792=117=0.05
Standard Normal Distribution Table
a. Calculate the test statistic.
z==
Round to two decimal places if necessary
Enter 0 if normal approximation to the binomial cannot be used
b. Determine the critical value(s) for the hypothesis test.
- +
Round to two decimal places if necessaryEnter 0 if normal approximation to the binomial cannot be used
c. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject
Cannot Use Normal Approximation to Binomial
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