null hypothesis p 0.90 against the alternative hypothesis p 0.90 at the 0.05 level of significance
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A: Givensample size(n)=1050x=95significance level(α)=0.05
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Q: A marriage counselor has traditionally seen that the proportion p of all married couples for…
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A: Solution: Given information: n= 200 Sample size of drivers x = 122 drivers drive an American made…
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A: given data claim : p > 0.231n = 200x = 52α = 0.05
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A: For the given data perform z test for one proportion
Q: A marriage counselor has traditionally seen that the proportion p of all married couples for whom…
A: Given data issample size(n)=205x=164sample proportion(p^)=xn=164205=0.8significance level(α)=0.05
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A: Given:x1=120, n1=250x2=132, n2=240 Then,p^1=x1n1=120250=0.48p^2=x2n2=132240=0.55…
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A: A survey of working adults indicates that only 50% of them are satisfied with their jobs. A…
Q: A marriage counselor has traditionally seen that the proportion p of all married couples for whom…
A: From the provided information,Sample size (n) = 210From which 167 of them stayed together.Level of…
Q: One research study of illegal drug use among 12- to 17-year-olds reported a decrease in use (from…
A: From the provided information, Sample size (n) = 1043 From which 96 report using illegal drugs.…
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Q: A marriage counselor has traditionally seen that the proportion p of all married couples for whom…
A:
Q: In a survey, 36% of the respondents stated that they talk to their pets on the telephone. A…
A: Sample size(n)=240No.of persons spoke to their pets on the telephone (x)=85Population proportion(p0…
Q: In a survey, 36% of the respondents stated that they talk to their pets on the telephone. A…
A: Sample size(n)=240No.of persons spoke to their pets on the telephone (x)=85Population proportion(p0…
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A: Given X1=20 X2=33 N1=345 N2=293
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A: The P-value is the probability of observing a sample proportion as extreme or more extreme than the…
Q: One study claimed that 86 % of college students identify themselves as procrastinators. A professor…
A: Answer Sample size [n] =139favorable cases [x] =110
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A: given data clim : p > 0.76n = 215x = 171α = 0.05p^ = xn = 171215 = 0.7953
Q: In a survey, 36% of the respondents stated that they talk to their pets on the telephone. A…
A: We have given thatp = 0.36Sample size (n) = 240x = 85Significance level () = 0.10
Q: A marriage counselor has traditionally seen that the proportion p of all married couples for whom…
A: Note:- Since you have posted a question with multiple subparts, we will provide the solution only to…
Q: One research study of illegal drug use among 12- to 17-year-olds reported a decrease in use (from…
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A: F-statistics are the ratio of two variances that are approximately the same value when the null…
Q: (a) State the null hypothesis Ho and the alternative hypothesis H₁. H:0 H₁:0 (b) Determine the type…
A: It is given that Population proportion, p = 80% = 0.80 Favourable cases, X = 178 Sample size, n =…
Q: One research study of illegal drug use among 12-to 17-year-olds reported a decrease in use (from…
A: The appropriate Null and Alternative Hypotheses are given below: From the given information, the…
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A: The level of significance is 0.01.
Q: A marriage counselor has traditionally seen that the proportion p of all married couples for whom…
A: Given that Sample size n =205 Favorable cases x =171 Sample proportion p^=x/n =171/205 =0.8341
A manufacturer of a spot remover claims that his product removes 90% of all spots. If, in
a sample, only 174 of 200 spots were removed with the manufacturer’s product, test the
null hypothesis p 0.90 against the alternative hypothesis p 0.90 at the 0.05 level of
significance
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- In the past, 30% of a country club's members brought guests to play golf sometime during the year. Last year, the club initiated a new program designed to encourage members to bring more guests to play golf. In a sample of 95 members, 34 brought guests to play golf after the program was initiated. When testing the hypothesis that the new program has increased the proportion of members bringing out guests (using a 5% level of significance), what is the null and alternative hypothesis?A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 77%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 77% of married couples. In a random sample of 250 married couples who completed her program, 194 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor's claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H₁. H₁:0 H₁ :0 (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) 0 (d) Find the p-value. (Round to three or more decimal places.) 0 (e) Is there enough…US Universities found that 72% of people are concerned about the possibility that their personal records could be stolen over the internet. If a random sample of 300 college students at a Midwestern university were taken and 228 of them were concerned about the possibility that their personal records could be stolen over the Internet, could you conclude at the 0.025 level of significance that a higher proportion of the university’s college students are concerned about Internet theft than the public at large? Report the p-value for this test. Z0.025 = 1.96
- A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 79%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 79% of married couples. In a random sample of 250 married couples who completed her program, 205 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor's claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H₁. H₂ : D H₁ : 0 (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) 0 (d) Find the p-value. (Round to three or more decimal places.) 0 (e) Is there enough…A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 77%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 77% of married couples. In a random sample of 215 married couples who completed her program, 167 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0 and the alternative hypothesis H1. H0: H1: (b) Determine the type of test statistic to use. ▼(Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the p-value. (Round to three…A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 78%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 78% of married couples. In a random sample of 240 married couples who completed her program, 189 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.10 level of significance?Perform a one-tailed test. Then complete the parts below.
- A fast-food restaurant manager wants to know if more than 60% of her customers that request a water cup actually fill the cup with soda instead of water. The manager would like to carry out a test at the a = 0.10 significance level of ho p=.60 ha p (greater than) .60 where p = the true proportion of customers that fill a water cup with soda rather than water. During the next month, the manager collected data from a random sample of 90 customers that ask for a water cup by observing that 63 of the customers filled the water cup with soda. Which of the following is the standardized test statistic for this hypothesis test? options 1.9365 2.0702 – 1.9365 – 2.0702One study claimed that 88 % of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 272 college students and finds that 231 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 88 % of college students are procrastinators? Use a 0.02 level of significance. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho p = 0.88 Ha P 0.88The Substance Abuse and Mental Health Services Administration (SAMHSA, 2017) estimates that 10.9% of the population of the United States age 18–24 had an episode of depression in the previous 12 months. Suppose a researcher takes a random sample of 225 United States adults aged 18–24. Determine the value of the sample proportion, p, such that 5% of samples of 225 United States adults age 18–24 are greater than p. You may find software or a z-table useful. Give your answer precise to three decimal places. p = II
- A publisher reports that79%of their readers own a personal computer. A marketing executivewants to test the claim that the percentage is actually more than the reported percentage. A randomsample of 100 found that89%of the readers owned a personal computer. Is there sufficient evidence atthe 0.02 level to support the executive’s claim?A nationwide survey of working adults indicates that out of 100 adults, 50 of them are satisfied with their jobs. The president of a large company believes that more than this number of employees at his company are satisfies with their jobs. To test his belief, he surveys a random sample of 100 employees, and 59 of them report that they are satisfied with their jobs. Do you support president's statement about the employee’s satisfaction? (Use a= 0.05 level of significance.) Find H0 and Ha. Find Test Statistic Identify Decision Identify ConclusionOne study claimed that 85 % of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 108 college students and finds that 86 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 85 % of college students are procrastinators? Use a 0.01 level of significance. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho :p = 0.85 0.85 d: "H Answer 囲 Tables 國 Keypad Keyboard Shortcuts O > b
