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- Please help with A,B,CQues45)The first and third quartile of a normal distribution are 51.70 and 55.30 respectively. Then the distribution is mode of theInternet Service Providers The amount that households pay service providers for access to the Internet varies quite a bit, but the mean monthly fee is $50 and the standard deviation is $20. The distribution is not Normal: many households pay a low rate as part of a bundle with phone or television service, but some pay much more for Internet only or for faster connections. A sample survey asks an SRS of 50 households with Internet access how much they pay. Let x be the mean amount paid. 1. Explain why you can't determine the probability that the amount a randomly selected household pays for access to the Internet exceeds $55. 2. What are the mean and standard deviation of the sampling distribution of x ? 3. What is the shape of the sampling distribution of x ? Justify your answer. 4. Find the probability that the average fee paid by the sample of households exceeds $55.
- dy The historical reports from two major networks showed that the mean number of commercials aired during prime time was equal for both networks last year. In order to find out whether they still air the same number of commercials on average or not, random and independent samples of 100 recent prime time airings from both networks have been considered. The first network aired a mean of 109.5 commercials during prime time with a standard deviation of 5.2. The second network aired a mean of 110.5 commercials during prime time with a standard deviation of 5.3. Since the sample sizes are quite large, assume that the population standard deviations can be estimated to be equal to the sample standard deviations, 5.2 and 5.3. At the 0.10 level of significance, is there sufficient evidence to support the claim that the mean number, μ₁, of commercials aired during prime time by the first station is not equal to the mean number, μ₂, of commercials aired during prime time by the second station?…On average, a banana will last 6.6 days from the time it is purchased in the store to the time it is too rotten to eat. Is the mean time to spoil different if the banana is hung from the ceiling? The data show results of an experiment with 14 bananas that are hung from the ceiling. Assume that that distribution of the population is normal. 7.9, 6, 5.5, 8.1, 7.9, 8.8, 6.6, 8.7, 6.2, 6.3, 8.6, 6.7, 8.5, 9.3 What can be concluded at the the α = 0.01 level of significance level of significance? a. For this study, we should use [t-test for a population mean b. The alternative hypothesis would be: H₁: Hv c. The test statistic d. Based on this, we should Select an answer e. Thus, the final conclusion is that ... Select an answer (please show your answer to 3 decimal places.) the null hypothesis.Shaded diagram.. 3. The random variable X has a normal distribution with mean 80 and standard deviation 6. Choose random samples of size 40 from this distribution. Let X be the random variable of sample means. (5) a. What is the distribution of X ? (5) b. What is the mean of X ? y
- Assume the samples are random and independent, the populations are nomally distributed, and the population variances are equal. The table available below shows the prices (in dollars) for a sample of automobile batteries. The prices are classified according to battery type. At a = 0.10, is there enough evidence conclude that at least one mean battery price is different from the others? Complete parts (a) through (e) below. E Click the icon to view the battery cost data. (a) Let u1. P2. H3 represent the mean prices for the group size 35, 65, and 24/24F respectively. Identify the claim and state Ho and H. H Cost of batteries by type The claim is the V hypothesis. Group size 35 Group size 65 Group size 24/24F 101 111 121 124 D 146 173 182 278 124 140 141 89 (b) Find the critical value, Fo, and identify the rejection region. 90 79 84 The rejection region is F Fo, where Fo = (Round to two decimal places as needed.) (c) Find the test statistic F. Print Done F= (Round to two decimal places as…a. Use a = 0.10 to test for a statistically significantly difference between the population means for first- and fourth-round scores. What is the p-value? t-value (to 3 decimals) Degrees of freedom p-value is greater than 0.20 What is your conclusion? There is no significant difference between the mean scores for the first and final rounds. b. What is the point estimate of the difference between the two population means? (to 2 decimals) For which round is the population mean score lower? First round c. What is the margin of error for a 90% confidence interval estimate for the difference between the population means? (to 3 decimals) Could this confidence interval have been used to test the hypothesis in part (a)? Yes Explain. Use the point of the difference between the two population means and add and subtract this MOE. If zero is in the interval the difference is not statistically significant. If zero is not in the interval the difference is statistically significant.The data in the table is points scored by a basketball team in a certain period of time. Use the data in the table to answer the bulleted statements.
- (a) Approximate the mean and standard deviation. Round each to two decimals. (b) Complete the last two columns in the table above for relative frequency and cumulative frequency. Round any decimals to 3 places. Do not include the zero before the decimal. For example for 0.56, enter .56 (c) What is the shape of the distribution?The accompanying histogram shows the lengths of hospital stays (in days) for all female patients admitted to a certain hospital during one year with a primary diagnosis of acute myocardial infarction (heart attack). Complete parts (a) through (c) below. a) From the histogram, determine whether the mean or median is larger. Explain. A. Because the distribution is nearly symmetric, the mean is expected to be larger than the median. B. Because the distribution is nearly symmetric, the mean is expected to be about the same as the median. C. Because the distribution is skewed to the high end, the mean is expected to be larger than the median. D. Because the distribution is skewed to the high end, the mean is expected to be smaller than the median. E. Because the distribution is skewed to the low end, the mean is expected to be larger than the median. F. Because the distribution is skewed to the low end, the mean is expected to be…