number with "1" as the leading digit is about 0.303. Now suppose you are the auditor for a very large corporation. The revenue file contains millions of numbers in a large computer data bank. You draw a random sample of n = 225 numbers from this file and r = 86 have a first nonzero digit of 1. Let p represent the population proportion of all numbers in the computer file that have a leading digit of 1. Test the claim that p is more than 0.303. Use ? = 0.10.   (a) What is the level of significance? State the null hypothesis  H0  and the alternate hypothesis  H1 . H0 : p          H1 : p          (b) What sampling distribution will you use? The Student's tThe standard normal     since np                and nq                What is the value of the sample test statistic? (Round your answer to two decimal places.) (c) Find the P-value of the test statistic. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value.         (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?? At the ? = 0.10 level, we reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.10 level, we reject the null hypothesis and conclude the data are not statistically significant.    At the ? = 0.10 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.10 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Interpret your conclusion in the context of the application. There is sufficient evidence at the 0.10 level to conclude that the true proportion of numbers with a leading 1 in the revenue file is greater than 0.303.There is insufficient evidence at the 0.10 level to conclude that the true proportion of numbers with a leading 1 in the revenue file is greater than 0.303.

A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Z-value p-value Single Policyholders n₁ = 440 Number making claims = 88 a. Use a = 0.05. Test to determine whether the claim rates differ between single and married male policyholders. (to 2 decimals) (to 4 decimals) X Married Policyholders n2 = 840 Number making claims 126 We can conclude ✔ that there is the difference between claim rates. b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number. X * )

A survey found that in a sample of 100 families, 37 owned cats. A survey done 5 years ago found that in a sample of 80 families, 34 owned cats. At ? = 0.05 has the proportion of cat owners changed over the 5-year period?

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 38 who smoke. Step 1 of 2:  Suppose a sample of 273 Americans over 38 is drawn. Of these people, 79 smoke. Using the data, estimate the proportion of Americans over 38 who smoke. Enter your answer as a fraction or a decimal number rounded to three decimal places.   Step 2 of 2:  Suppose a sample of 273 Americans over 38 is drawn. Of these people, 79 smoke. Using the data, construct the 85% confidence interval for the population proportion of Americans over 38 who smoke. Round your answers to three decimal places.   Lower end point, upper end point

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The FBI wants to determine the effectiveness of their 10 most wanted list. To do so, they need to find out the fraction of people who appear on the list that are actually called. Suppose it suppose a sample of 531 suspected criminals is drawn period of these people, 148 were captured. Using the data, estimate the proportion of people who were caught after being on the 10 most wanted list. Enter your answer as a fraction or a decimal number round it to 3 decimal places.

A student pursuing a degree in English as a second language believes the proportion female factory workers who can't speak English is less than the proportion of male factory workers who can't speak English. To test her claim she randomly selects 373 female factory workers and out of them 43 could not speak English. She then randomly selects 215 male factory workers and out of them 65 could not speak English. Test her claim at .=0.05 to see if she was right. The correct hypotheses are: H : PF Pu (claim) H : PF 2 PM HA: PF < Pr (claim) H : PF = PM HA: PF + PM (claim) Since the level of signficance s0.05 the critical value is-1.64s The test statistic is pound to 3 places) The pvalue is round to 3 places) The decision can be made to oreject H, odo not reject í, The final conclusion is that: There is enough evidence to reject the claim that the proportion female factory workers who can't speak English is less than the proportion of male factory workers who can't speak English. There is…

How would I identify the p-value?