find the 95% confidence interval estimate of the population mean
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Use the given data to find the 95% confidence
IQ scores of professional athletes:
Mean = 104
Standard deviation = 9
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- The average monthly, entry-level salary for fresh graduates was reported as $2,800. A recruiting agency believed that the mean salary of fresh entry is lower than this. The recruiting office randomly surveyed 25 entry-level graduates and found that the sample means salary is $2,200, and the sample standard deviation is $1800. Assuming t- distribution is followed, answer the following questions. Use: H0 : u>2800 H1 : uConstruct a confidence interval for the population mean. Assume that the population has a normal distribution. n= 30, x bar = 84.6, s=10.5, 90% confidence.Normal Distribution. The volume of water in commercially supplied fresh drinking water containers is approximately Normally distributed with mean 70 liters and standard deviation 0.75 liters. Estimate the proportion of containers likely to contain less than 70.5 liters. Select the correct response: 0.0082 0.8200 0.7468 0.1151
- Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of error and 95% confidence interval. n=49, x=55.1 seconds, s=4.8 seconds The margin of error is nothing seconds.Solve the problem and draw the normal curve with the shaded region representing the given scenario in the problem. Explicitly write your solution and state & highlight your final answer at the end. You have just received your score in a mathematics aptitude test. Given your score was 82 and the mean score on the test was 73 with a standard deviation of 5. Assume that the test scores are normally distributed. a. Find the median and the mode, and compare it with the given mean value. b. Determine the intervals within 1, 2 and 3 standard deviations from the mean, that is, the µ ± 10, µ ± 20, μ ± 30.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 to 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 u. 2 Ha represent the mean prices for the group size 35, 65, and 24/24F respectively. Identify the claim and state Ho and H.. Ho: H - X Cost of batteries by type The claim the V hypothesis. Group size 35 Group size 65 Group size 24/ 24F 91 99 111 119 126 O 91 144 174 180 278 79 84 126 140 140 (b) Find the critical value, Fo. and identify the rejection region. The rejection region is F Fo, where Fo- (Round to two decimal places as needed.) (c) Find the test statistic F. F= (Round to two decimal places as…
- Given a mean, standard deviation, and a raw score, find the corresponding z-score. Assume the distribution is normal. mean 70, standard deviation 4, x = 80 What is the corresponding z-score? z=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…The distribution of heights in a population of women is approximately normal. Sixteen percent of the women have heights less than 62 inches. About 97.5% of the women have heights less than 71 inches. Use the empirical rule to estimate the mean and standard deviation of the heights in this population. Mean: K inches Standard Deviation: inches
- For the population of one town, the number of siblings, x, is a random variable whose relative frequency histogram has a reverse J-shape. The mean number of siblings is 1.3 and the standard deviation is 1.5. Let x denote the mean number of siblings for a random sample of size 32. Determine whether the distribution of x is normal or approximately normal and find its mean and standard deviation.Use z-scores to make the following comparison. A high school student took two college entrance exams, scoring 1110 on the SAT and 27 on the ACT. Suppose that SAT scores have a mean of 950 and a standard deviation of 130 while the ACT scores have a mean of 22 and a standard deviation of 4. Assuming the performance on both tests follows a normal distribution, determine which test the student did better on.Use the normal distribution of IQ scores , which has a mean of 90 and a standard deviation of 14, and the following table with the standard scores and percentiles for a normal distribution to find the indicated quantity. view the table: Standard Scores and Percentiles for a Normal Distribution (cumulative values from the left) Full data set Open in StatCrunch + Copy to Clipboard + Open in Excel + Standard score % Standard score % minus−3.0 0.13 0.1 53.98 minus−2.5 0.62 0.5 69.15 minus−2 2.28 0.9 81.59 minus−1.5 6.68 1 84.13 minus−1 15.87 1.5 93.32 minus−0.9 18.41 2 97.72 minus−0.5 30.85 2.5 99.38 minus−0.1 46.02 3 99.87 0 50.00 3.5 99.98 The percentage of scores between 62 and 118 is ________%. Round to two decimal places as needed.
