n = 49 HỌ: µ = 50 x= 54.8 Ha: µ ± 50 O = 28 0. Refer to Exhibit 9-3. If the test is done at a 5% level of significance, the null ypothesis should not be rejected be rejected be tested again Not enough information is given to answer this question.

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Aspirin is packed in 120 tab bottles with weight normally distributed with mean 600g and standard deviation 6g.  Rejected bottles are those whose weight differs from the mean weight by 15g or more.  That is, rejected bottles have weight either too small or too large.  We want to find out how many bottles will be rejected from a daily production of 5000 bottles. 1,Do we have to calculate one z-score or two z-scores?     2. The negative z-score is    3. The positive z-score is      4.How many bottles, of the 5000, do we anticipate to have to reject?

on 5 of 21 The heights of women aged 20-29 in the United States are approximately Normal with mean 64.1 inches and standard deviation 3.7 inches. Men the same age have mean height 69.4 inches, with standard deviation 3.1 inches. Cheryl D. Fryar et al., "Anthropometric reference data for children and adults: United States, 2011-2014," Vital and Health Statistics, Series 3, no. 39 (August 2016), at www.cdc.gov/nchs. This report provides the means of various anthropometric measurements. Standard deviations were computed from the sample size and reported standard errors of the mean. Instructions on measuring upper arm length are in the National Health and Nutrition Examination Survey: Anthropometry Procedures Manual, January 2007. Side-by-Side O Macmillan Learning What is the z-score for a woman 5.5 feet tall? Give your answer to two decimal places. Z =

Professor Ramos has a packing plant in a machine packs carton with jars. A sales person claims that the machine she is selling will pack faster to test the claim the times It takes each machine to pack 10 cartons are recorded in seconds and the results are shown below population one new machine n=10 Mean =42.3 s=0.685 Present machine n=10 mean=43.23 S =0.750.Do the data provide significant evidence to conclude that on the average the new machine packs faster perform the required hypothesis test in the 5% level of significance assume the population standard deviation's are equal but unknown show all your steps setting up your null  and alternative hypothesis.

In a poll, 1000 adults in a region were asked about their online vs. in-store clothes shopping. One finding was that 37% of respondents never clothes-shop online. Find and interpret a 95% confidence interval for the proportion of all adults in the region who never clothes-shop online. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. The 95% confidence interval is from to. (Round to three decimal places as needed.) Interpret the confidence interval. Choose the correct answer below. A. There is a 95% probability that the sample proportion of a survey that asks adults in the region whether they clothes-shop online will be within the bounds of the confidence interval found in the previous step. B. With 95% confidence, the true percentage of all adults in the region who never clothes-shop online is within the bounds of the confidence interval found in the previous step. C. With…

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