If a binomial distribution satisfies the criterion for using the normal approximation with p = 0.15, then what is the lowest value of n that still meets the sample size assumptions of the binomial test?​ Select one: a. ​ n = 100 b. ​ n = 150 c. ​ n = 81 d. ​ n = 67

A poll of 1136 teens aged 13 to 17 showed that 53% of them have made new friends online. Use a 0.10 significance level to test the claim that half of all teens have made new friends online. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. Let p denote the population proportion of all teens aged 13 to 17 who have made new friends online. Identify the null and alternative hypotheses. Ho: P H₁: p (Type integers or decimals. Do not round.) Identify the test statistic. Z= (Round to two decimal places as needed.) Identify the P-value. C P-value= (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Reject the null hypothesis. There sufficient evidence to the claim that half of all teens have made new friends online.

The sample size n, probability of success p, and probability of failure q are given for a binomial experiment. Decide whether you can use the normal distribution to approximate the random variable x. n = 19 q=0.25 0000 Can the normal distribution be used to approximate the random variable x? A. No, because nq < 5. B. No, because np < 5 and nq < 5. p=0.75 OC. No, because np < 5. O D. Yes, because np 25 and nq ≥ 5. www.

A poll of 1004 teens aged 13 to 17 showed that 57% of them have made new friends online. Use a 0.01 significance level to test the claim that half of all teens have made new friends online. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. C Let p denote the population proportion of all teens aged 13 to 17 who have made new friends online. Identify the null and alternative hypotheses. Ho: P H₁: p (Type integers or decimals. Do not round.) Identify the test statistic. 2=0 (Round to two decimal places as needed.) Identify the P-value. P-value= (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. the null hypothesis. There ▼sufficient evidence to the claim that half of all teens have made new friends online.

A poll of 2,115 randomly selected adults showed that 87% of them own cell phones. The technology display below results from a test of the claim that 94% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.94 vs p *0.94 Sample X N 1 1844 2,115 Sample p 0.871868 95% CI (0.853147,0.890588) a. Is the test two-tailed, left-tailed, or right-tailed? Z-Value - 13.19 P-Value 0.000

The president of Doerman Distributors, Inc., believes that 31% of the firm's orders come from first-time customers. A simple random sample of 100  orders will be used to estimate the proportion of first-time customers. Use z-table. a. Assume that the president is correct and p=.31 What is the sampling distribution of ˉp   for this study? - Select your answer A non normal distribution A normal distribution because np and n(1-p) are both greater than 5 A normal distribution because np and n(1-p) are both less than 5Item 1 b.What is the probability that the sample proportion will be between .22. and  .40 (to 4 decimals)?   c. What is the probability that the sample proportion will be between .27 and .35 (to 4 decimals)?

A poll of 2,153 randomly selected adults showed that 88​% of them own cell phones. The technology display below results from a test of the claim that 90​% of adults own cell phones. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.05 significance level to complete parts​ (a) through​ (e).   Test pf p= 0.9 vs p≠ 0.9   Sample 1 X 1893  N 2,153  Sample p 0.879238 95% CI (0.865474, 0.893002) Z-Value -3.21 P-Value 0.001   Is it a two tailed?Right tailed? Left tailed? What is the test statistic? What is the P-value? What is the null hypothesis and what is concluded about it? What is the final conclusion?

A poll of 1172 teens aged 13 to 17 showed that 55% of them have made new friends online. Use a 0.05 significance level to test the claim that half of all teens have made new friends online. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. Let p denote the population proportion of all teens aged 13 to 17 who have made new friends online. Identify the null and alternative hypotheses. Ho: P H₁: P (Type integers or decimals. Do not round.) Identify the test statistic.. z= (Round to two decimal places as needed.) Identify the P-value. P-value= (Round to three decimal places as needed.) = State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. the null hypothesis. There sufficient evidence to Accept Reject Fail to reject CID is is not the claim that half of all teens have made new friends a support warrant rejection of