Question 13 2 pts (CO6) Suppli
Question 13 2 pts
(CO6) Supplier claims that they are 95% confident that theirproducts will be in the interval of 20.45 to 21.05. You takesamples and find that the 95% confidence interval of what they aresending is 20.50 to 21.10. What conclusion can be made?
| The supplier is more accurate than they claimed |
| The supplier products have a lower mean than claimed |
| The supplier products have a higher mean than claimed |
| The supplier is less accurate than they have claimed |
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Question 14 2 pts
(CO6) In a sample of 17 small candles, the weight is found to be3.72 ounces with a standard deviation of 0.963 ounces. What wouldbe the 87% confidence interval for the size of the candles?
| (3.369, 4.071) |
| (3.199, 4.241) |
| (3.347, 4.093) |
| (2.757, 4.683) |
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Question 15 2 pts
(CO6) In a situation where the sample size was increased from 29to 39, what would be the impact on the confidence interval?
| It would become wider due to using the t distribution |
| It would become narrower with fewer values |
| It would remain the same as sample size does not impactconfidence intervals |
| It would become narrower due to using the z distribution |
Question 23 2 pts
(CO7) A light bulb manufacturer guarantees that the mean life ofa certain type of light bulb is at least 720 hours. A random sampleof 51 light bulbs as a mean of 705.4 hours with a standarddeviation of 62 hours. At an α=0.05, can you support the company’sclaim using the test statistic?
| Claim is the null, reject the null and cannot support claim astest statistic (-1.68) is in the rejection region defined by thecritical value (-1.645) |
| Claim is the alternative, fail to reject the null and cannotsupport claim as the test statistic (-1.68) is in the rejectionregion defined by the critical value (-1.96) |
| Claim is the alternative, reject the null and support claim astest statistic (-1.68) is not in the rejection region defined bythe critical value (-1.96) |
| Claim is the null, fail to reject the null and cannot supportclaim as test statistic (-1.68) is not in the rejection regiondefined by the critical value (-1.645) |
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Question 24 2 pts
(CO7) A restaurant claims the customers receive their food inless than 16 minutes. A random sample of 39 customers finds a meanwait time for food to be 15.8 minutes with a standard deviation of0.7 minutes. At α = 0.04, can you support the organizations’ claimusing the test statistic?
| Claim is the alternative, reject the null so support the claimas test statistic (-1.78) is in the rejection region defined by thecritical value (-1.75) |
| Claim is the null, reject the null so cannot support the claimas test statistic (-1.78) is in the rejection region defined by thecritical value (-2.05) |
| Claim is the alternative, fail to reject the null so cannotsupport the claim as test statistic (-1.78) is not in the rejectionregion defined by the critical value (-2.05) |
| Claim is the null, fail to reject the null so support the claimas test statistic (-1.78) is not in the rejection region defined bythe critical value (-1.75) |
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Question 25 2 pts
(CO7) A manufacturer claims that their calculators are 6.800inches long. A random sample of 55 of their calculators finds theyhave a mean of 6.812 inches with a standard deviation of 0.05inches. At α=0.08, can you support the manufacturer’s claim usingthe p value?
| Claim is the alternative, fail to reject the null and supportclaim as p-value (0.075) is less than alpha (0.08) |
| Claim is the null, fail to reject the null and support claim asp-value (0.038) is greater than alpha (0.08) |
| Claim is the alternative, reject the null and cannot supportclaim as p-value (0.038) is greater than alpha (0.08) |
| Claim is the null, reject the null and cannot support claim asp-value (0.075) is less than alpha (0.08) |
Answer:
Solution:-
25)
Claim is the null, reject the null and cannot supportclaim as p-value (0.075) is less than alpha (0.08).
State the hypotheses. The first step is tostate the null hypothesis and an alternative hypothesis.
Null hypothesis: u = 6.800Alternative hypothesis: u 
6.800
Note that these hypotheses constitute a two-tailed test. Thenull hypothesis will be rejected if the sample mean is too big orif it is too small.
Formulate an analysis plan. For this analysis,the significance level is 0.05. The test method is a one-samplet-test.
Analyze sample data. Using sample data, wecompute the standard error (SE), degrees of freedom (DF), and the tstatistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.00674
DF = n – 1
D.F = 54t = (x – u) / SE
t = 1.78
where s is the standard deviation of the sample, x is the samplemean, u is the hypothesized population mean, and n is the samplesize.
Since we have a two-tailed test, the P-value is the probabilitythat the t statistic having 54 degrees of freedom is less than-1.78 or greater than 1.78.
Thus, the P-value = 0.0751.
Interpret results. Since the P-value (0.0751) is lessthan the significance level (0.08), we have to reject the nullhypothesis.
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