golf project

. State the interval estimate that contains more possible values of the population mean (precision) for a 90% CI or a 95% CI for the same data.

3) both photos are the same question

What is the approximate forecast for May using a four-month moving average? Nov. Dec. Jan. Feb. Mar. April. 39 OA. 38 36 40 42 48 46 O B. 44 OC.43 O D. 42

Consider a set of numbers 25.8, 30, 21.9, 20.3, 29.5, 21.2,27.3, 20.8 and 27.1. Find the IQR of this data, to 2 decimal places.

Please answer the following question by stating conditions and test required. Please show how to answer using graphing calculator if possible.

Data from the Office for National Statistics show that the mean age at which men in Great Britain get married was 33.5. A news reporter noted that this represents a continuation of the trend of waiting until a later age to wed. A new sample of 47 recently wed British men provided their age at the time of marriage. These data are contained in the Excel Online file below. Construct a spreadsheet to answer the following questions. Open spreadsheet Do these data indicate that the mean age of British men at the time of marriage exceeds the mean age in 2013? Test this hypothesis at a = 0.05. What is your conclusion? Use the obtained rounded values in your calculations. Sample mean: Sample standard deviation: t-value: p-value (Upper Tail): Because p-value > ✓ years (to 2 decimals) years (to 4 decimals) (to 3 decimals) (to 3 decimals) a = 0.05, we fail to reject ✔✔✔ Ho. There is insufficient ✔✔✔ evidence to conclude that the mean age at which British men get married exceeds what it was in…

please help these answers are wrong and 0.92, 0.27, and 15 are not correct either

The average American gets a haircut every 39 days. Do college students get their hair cut less frequently? The data below shows the results of a survey of college students asking them how many days elapse between haircuts. Assume that the distribution of the population is normal. 40, 43, 57, 60, 39, 35, 51, 23, 30, 43, 56, 48, 36, 31, 44, 50, 60 What can be concluded at the 0.05 level of significance? Helpful Videos:CalculationsLinks to an external site., Set-upLinks to an external site., InterpretationsLinks to an external site. HintLinks to an external site. Textbook PagesLinks to an external site. H0: mu.gif = 39 Ha: mu.gif [ Select ] 39 Test statistic: [ Select ] p-Value = [ Select ] [ Select ] Conclusion: There is sufficient evidence to make the conclusion that the population mean number of days between haircuts for college students is more than 39.

0.7200.7400.6400.3900.7002.2001.9800.6401.2200.2001.6401.3402.9500.9001.7601.0101.2600.0000.6501.4601.6201.8300.9901.5600.4101.2800.8301.3200.5401.2500.9201.0000.7800.7901.4401.0002.2402.5001.7901.2501.4900.8401.4201.0001.2501.4201.3500.9300.4001.390 The accompanying data table lists the magnitudes of 50 earthquakes measured on the Richter scale. Test the claim that the population of earthquakes has a mean magnitude greater than 1.00. Use a 0.01significance level. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, and conclusion for the test. Assume this is a simple random sample the sample data is above.   What are the​ hypotheses?     A. H0​: μ=1.00 in magnitude H1​: μ≠1.00 in magnitude   B. H0​: μ=1.00 in magnitude H1​: μ<1.00 in magnitude   C. H0​: μ=1.00 in magnitude H1​: μ>1.00 in magnitude   D. H0​: μ≠1.00 in magnitude H1​: μ=1.00 in magnitude   Identify the test statistic.   t= ​(Round to two decimal places as​…

Use the sample data and confidence level given below to complete parts (a) through (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n= 1048 and x = 568 who said "yes." Use a 95% confidence level. Click the icon to view a table of z scores. Standard Normal (z) Distribution a) Find the best point estimate of the population proportion p. .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 (Round to three decimal places as needed.) 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.0 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.1 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.2 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.3 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.4 0.5 .6915 .6950 .6985 7019 .7054 .7088 .7123 .7157 .7190 .7224 0.5 0.6 .7257 .7291 7324 7357 .7389 .7422 .7454 .7486 .7517 .7549 0.6 0.7 .7580 .7611 .7642 7673 7704 .7734 .7764 .7794…

39. The following table gives the approximate increase in sea level in inches over 20 years starting in the given year. Estimate the net change in mean sea level from 1870 to 2010. Starting Year 1870 1890 1910 1930 1950 1970 1990 20-Year Change 0.3 1.5 0.2 2.8 0.7 1.5 4 Table 5.3 Approximate 20-Year Sea Level Increases, 1870-1990 Source: http://link.springer.com/article/ 10.1007%2Fs10712-011-9119-1

Please only do B, C, and D...

The average American gets a haircut every 44 days. Is the average smaller for college students? The data below shows the results of a survey of 20 college students asking them how many days elapse between haircuts. Assume that the distribution of the population is normal. 43, 44, 49, 47, 32, 38, 30, 40, 45, 48, 34, 37, 48, 37, 35, 42, 42, 31, 40, 50 Copy to clipboard What can be concluded at the = 0.10 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Но: Select an answer ? v Select an answer v c. The test statistic ? v (please show your answer to 3 decimal places.)