# 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 = 54**t = (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.**