significance test
A brand of soda is supposed to contain 12 oz. A quality control inspector has been hired to find out if the actual can fill differs from this, and selects 8 cans at random to find an average of 12.03 oz. with standard deviation 0.073 oz. Test the claim at the 0.05 level of significance.
H0 : μ = 12
H1 : μ ≠ 12
a) enter the critical t value from the t-Distribution table (3 decimal places).
b) compute the sample t
t=x⎯⎯⎯−μsn√ j
c) enter the values the sample t is between in the table
(Ex. if df = 16 and the sample t is 1.256, you would enter: 1.071<1.256<1.337)
d) enter the p-value estimate
(Ex. if df = 16 and 2.235<2.251<2.583, the areas at the top of the table would be 0.02 and 0.01, you would enter 0.01<p<0.02 for a one-tailed estimate and 0.02<p<0.04 for a two-tailed estimate)
Homework Answers
USE THIS TABLE
Give as much information as you can about the P-value of a t test in each...
Give as much information as you can about the P-value of a t test in each of the following situations. (Round your answers to three decimal places.) (a) Upper-tailed test, df = 9, t = 2.0 P-value < 0.005 0.005 < P-value < 0.01 0.01 < P-value < 0.025 0.025 < P-value < 0.05 P-value > 0.05 (b) Upper-tailed test, n = 13, t = 3.2 P-value < 0.005 0.005 < P-value < 0.01 0.01 < P-value < 0.025 0.025...
Test the claim that the mean GPA of night students is smaller than 2.9 at the...
Test the claim that the mean GPA of night students is smaller than 2.9 at the 0.01 significance level. The null and alternative hypothesis would be: H0:p≥0.725H0:p≥0.725 H1:p<0.725H1:p<0.725 H0:μ≥2.9H0:μ≥2.9 H1:μ<2.9H1:μ<2.9 H0:μ=2.9H0:μ=2.9 H1:μ≠2.9H1:μ≠2.9 H0:μ≤2.9H0:μ≤2.9 H1:μ>2.9H1:μ>2.9 H0:p=0.725H0:p=0.725 H1:p≠0.725H1:p≠0.725 H0:p≤0.725H0:p≤0.725 H1:p>0.725H1:p>0.725 The test is: right-tailed two-tailed left-tailed Based on a sample of 65 people, the sample mean GPA was 2.89 with a standard deviation of 0.02 The p-value is: (to 2 decimals) Based on this we: Fail to reject the null hypothesis Reject the...
You are conducting a significance test of H0: μ = 5 against Ha: μ > 5....
You are conducting a significance test of H0: μ = 5 against Ha: μ > 5. After checking the conditions are met from a simple random sample of 30 observations, you obtain t = 2.35. Based on this result, describe the p-value. The p-value falls between 0.15 and 0.2. The p-value falls between 0.025 and 0.05. The p-value falls between 0.01 and 0.02. The p-value falls between 0.005 and 0.01. The p-value is less than 0.005.
Consider tlowing hypotheses: H0: μ-12 HA12 Find the p-value for this test based on the following...
Consider tlowing hypotheses: H0: μ-12 HA12 Find the p-value for this test based on the following sample information. (You may find it useful to reference the appropriate table: z table or t table) a. X-11:s-3.2; n-36 Op-value<0.01 0.01 s p-value0.02 0.02 s p-value0.05 0.05 s p-value<0.10 O p-value 2 0.10 b. X = 13, s= 3.2; n= 36 p-value <0.01 0.01 s p-value0.02 0.02 s p-value<0.05 0.05 s p-value<0.10 p-value 20.10 C. X = 11, s = 2.8; n =...
Find the critical value(s) and rejection region(s) for the indicated t test level of significance α...
Find the critical value(s) and rejection region(s) for the indicated t test level of significance α and sample size n Left-tailed test, α: 0.005, n 7 Click the icon to view the t-distribution table. The critical value(s) isare (Round to the nearest thousandth as needed. Use a comma to separate answers as needed ) Determine the rejection region(s) Select the correct choice below and filt in the answer boxies) within your choice Round to the nearest thousandth as needed) OB...
5. You are given the following information about a hypothesis test that is being performed on...
5. You are given the following information about a hypothesis test that is being performed on a random sample taken from a normal distribution with mean ji: • The unbiased sample variance is 9. • The sample size is 16. • H,:= 80 • Ha:> 80 Let tay be the critical value of a t random variable with v degrees of freedom. The following table lists values of ta, for specific combinations of a and v: v = 15 v...
A 0.01 significance level is used for a hypothesis test of the claim that when parents...
A 0.01 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is 5 less than 0.5. Assume that sample data consists of 55 girls in 121 births, so the sample statistic of results in a z score that is 1 standard deviation below 0. Complete parts (a) through (h) below. Click here to view page 1 of the Normal table. Click here to...
Consider the following hypothesis test. H0: μ ≤ 12 Ha: μ > 12 A sample of 25 provided a sample mean x = 14 and a sample standard deviation s = 4.32. (a) Compute the value of the test statistic. (Ro...
Consider the following hypothesis test. H0: μ ≤ 12 Ha: μ > 12 A sample of 25 provided a sample mean x = 14 and a sample standard deviation s = 4.32. (a) Compute the value of the test statistic. (Round your answer to three decimal places.) _______ (b) Use the t distribution table to compute a range for the p-value. a) p-value > 0.200 b) 0.100 < p-value < 0.200 c) 0.050 < p-value < 0.100 d) 0.025 <...
The price to earnings ratio (P/E) is an important tool in financial work. A random sample...
The price to earnings ratio (P/E) is an important tool in
financial work. A random sample of 14 large U.S. banks (J. P.
Morgan, Bank of America, and others) gave the following P/E
ratios.†
24
16
22
14
12
13
17
22
15
19
23
13
11
18
The sample mean is
x ≈ 17.1.
Generally speaking, a low P/E ratio indicates a "value" or
bargain stock. Suppose a recent copy of a magazine indicated that
the P/E ratio of...
Consider the following hypothesis test. H0: μ ≤ 12 Ha: μ > 12 A sample of...
Consider the following hypothesis test. H0: μ ≤ 12 Ha: μ > 12 A sample of 25 provided a sample mean x = 14 and a sample standard deviation s = 4.28. (a) Compute the value of the test statistic. (Round your answer to three decimal places.) (b) Use the t distribution table to compute a range for the p-value. p-value > 0.2000.100 < p-value < 0.200 0.050 < p-value < 0.1000.025 < p-value < 0.0500.010 < p-value < 0.025p-value <...
Consider tlowing hypotheses: H0: μ-12 HA12 Find the p-value for this test based on the following sample information. (You may find it useful to reference the appropriate table: z table or t table) a. X-11:s-3.2; n-36 Op-value<0.01 0.01 s p-value0.02 0.02 s p-value0.05 0.05 s p-value<0.10 O p-value 2 0.10 b. X = 13, s= 3.2; n= 36 p-value <0.01 0.01 s p-value0.02 0.02 s p-value<0.05 0.05 s p-value<0.10 p-value 20.10 C. X = 11, s = 2.8; n =...
Find the critical value(s) and rejection region(s) for the indicated t test level of significance α and sample size n Left-tailed test, α: 0.005, n 7 Click the icon to view the t-distribution table. The critical value(s) isare (Round to the nearest thousandth as needed. Use a comma to separate answers as needed ) Determine the rejection region(s) Select the correct choice below and filt in the answer boxies) within your choice Round to the nearest thousandth as needed) OB...
5. You are given the following information about a hypothesis test that is being performed on a random sample taken from a normal distribution with mean ji: • The unbiased sample variance is 9. • The sample size is 16. • H,:= 80 • Ha:> 80 Let tay be the critical value of a t random variable with v degrees of freedom. The following table lists values of ta, for specific combinations of a and v: v = 15 v...
A 0.01 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is 5 less than 0.5. Assume that sample data consists of 55 girls in 121 births, so the sample statistic of results in a z score that is 1 standard deviation below 0. Complete parts (a) through (h) below. Click here to view page 1 of the Normal table. Click here to...
The price to earnings ratio (P/E) is an important tool in
financial work. A random sample of 14 large U.S. banks (J. P.
Morgan, Bank of America, and others) gave the following P/E
ratios.†
24
16
22
14
12
13
17
22
15
19
23
13
11
18
The sample mean is
x ≈ 17.1.
Generally speaking, a low P/E ratio indicates a "value" or
bargain stock. Suppose a recent copy of a magazine indicated that
the P/E ratio of...
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