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✏ Practice Exam 2a (42 Questions)
Due No due date Points 42 Questions 42 Time Limit 75 Minutes
Allowed Attempts Unlimited
Attempt History
Attempt Time Score LATEST Attempt 1 75 minutes 15 out of 42 Submitted Apr 22 at 10:26pm Take the Quiz Again The following 5 questions are based on this information. Suppose a random sample of 18 uber drivers was taken, and the average age x̄ of these drivers was found to be 27.5 years. Construct a 99% confidence interval for the mean age (μ) of all drivers. Assume that the age of drivers is normally distributed with a population standard deviation (σ) of 2.3 years. Question 1^1 / 1^ pts The standard error (SE) of x̄ is
Correct!Correct! 0.
Your answer is correct. SE= sigma sqrt (n) n 18 mean 27.5 sigma 2. Question 2^0 / 1^ pts The critical value (CV) used for a 99% interval estimate is
orrect Answerorrect Answer 2. ou Answeredou Answered 0.
Your answer is incorrect. CV = -NORM.S.INV(0.01/2) Question 3^1 / 1^ pts The 99% confidence interval estimate of μ is
Question 5^0 / 1^ pts If we decrease the confidence level (1-α) from 0.99 to 0.95, the margin of error (ME) of the confidence interval estimate will ou Answeredou Answered be zero stays the same decrease orrect Answerorrect Answer increase Your answer is incorrect. ME is decreases with decrease in the confidence level. The following 6 questions are based on this information. A large cycle company reported that in 2010, 65% of parents of children (between the ages of 8 to 12), bought their children a cycle. That same company has stated that cycle purchases for that same age group have gone up. They surveyed 300 parents of children and found that 230 of them have bought cycle to their children. Question 6^0 / 1^ pts Specify the null and alternative hypotheses.
orrect Answerorrect Answer H(0): p≤0.65 Versus H(a): p>0. H(0): p≥0.65 Versus H(a): p<0. ou Answeredou Answered That same company has stated that cell phone purchases for that same age group have gone up. Hence, H(0): p≤0.65 Versus H(a): p>0. Question 7^1 / 1^ pts The standard error (SE) of p¯ is Correct!Correct! 0.
SE= =SQRT(p*(1-p)/n)) p = 0. n = 300 The following 5 questions are based on this information.
orrect Answerorrect Answer
Your answer is incorrect. mean 3000 n 60 s 600 SE 77.45 SE= s/ sqrt(n) Question 10^0 / 1^ pts The test statistics value is orrect Answerorrect Answer 6. ou Answeredou Answered 0.
Your answer is incorrect. mean 3000 x 3500 n 60 s 600 t obs 3.10 SE 64. T obs = x - mean SE
Question 11^0 / 1^ pts The p-value is
ou Answeredou Answered 0.
orrect Answerorrect Answer 0. Your answer is incorrect. This is right-tailed test. P value= =1-T.DIST(tobs,n-1,TRUE) Question 12^0 / 1^ pts At α=0.01 and using the p-value We reject H(0) in favor of H(a) orrect Answerorrect Answer ou Answeredou Answered We do not reject H(0) Your answer is incorrect. if p value
ou Answeredou Answered 10.
orrect Answerorrect Answer 0. Your answer is incorrect. SE = sigma sqrt(n) n 30 x bar 10 sigma 4. Question 15^1 / 1^ pts The p-value is
Correct!Correct! 0.
Your answer is correct. p value = NORM.S.DIST(Z obs,TRUE)
Question 16^0 / 1^ pts The test statistics value is -2. orrect Answerorrect Answer ou Answeredou Answered 0.
-2. Your answer is incorrect. n 30 x bar 10 sigma 4.5 mean 0 12 SE 0.9 z obs -2. Z obs = x bar - mean 0 SE Question 17^1 / 1^ pts At α=0.10 and using the p-value We reject H(0) in favor of H(a) Correct!Correct! We do not reject H(0)
Question 19^0 / 1^ pts The critical value (CV) needed for 99% confidence interval estimation is orrect Answerorrect Answer 2.
ou Answeredou Answered 1. Your answer is incorrect. P value =-NORM.S.INV(0.01/2) Question 20^0 / 1^ pts The 99% confidence interval estimate of p is 0.75 ± 0. orrect Answerorrect Answer 0.75 ± 0. ou Answeredou Answered 0.44 ± 0. 0.44 ± 0.
Your answer is incorrect. p bar 0.75 n 500 alpha 0.01 SE 0.019365 CV 2.575829 ME 0.049881 ME = CV*SE IE= pbar +/- ME = 0.75 +/- 0. Question 21^1 / 1^ pts Suppose around the period the above poll was conducted,The DEan of a university made a personal statement saying that .85% of Graduate students used only the Internet for assignment purposes In light of the sample evidence and at the 1% level of significance, Correct!Correct! We can reject the Dean's claim We cannot reject the Dean's claim Your answer is correct. Because 0.85 is not in the range 0.75 +/- 0.05. We can reject the claim. Question 22^0 / 1^ pts A Dean of the universityl wishes to collect new random sample with the aim of building a new confidence interval at the 99% confidence level for p. Using the current sample proportion (from the 500 graduate students poll ) as a basis, what sample size (n) would the journalist require to achieve a
orrect Answerorrect Answer is normal because np≥5 and n(1−p)≥ ou Answeredou Answered is not normal because the sample size is too small is not normal because n < 500 is normal because the only requirement is for n to be greater than 30 and that is met Your answer is incorrect. The sampling distribution of p ̅ is Normal, when np>=5 and n(1- p)>= Question 24^0 / 1^ pts The standard error (SE) of p¯ is
ou Answeredou Answered
orrect Answerorrect Answer 0.
Your answer is incorrect. σ p ̅ = √((p(1-p))/n) p= 0. n= 400 Question 25^1 / 1^ pts What is the probability that a random sample of 400 U.S. youth will provide a sample proportion (p¯) that is within 0.03 of the population proportion (p)? Correct!Correct! 76. 57% 23% 43% Your answer is correct. n 400 n 400 mean 0.51 mean 0.51 x1 bar 0.54 x2 bar 0.48 Z 1.20024005 Z -1.20024005 SE 0.024995 SE 0.024995 Probablity 77% Hint: within 0.03 of the population proportion Question 26^0 / 1^ pts
This is a rare finding because the likelihood of \p¯=0.42 is quite small as we saw in the previous question orrect Answerorrect Answer Your answer is incorrect. 0.42 is out of range of. x1 bar 0.58 x2 bar 0. The following 6 questions are based on this information. According to a journalist of BBC News from April 2016, the average price of milk at all U.S. supermarkets was $3.00 (μ). The population standard deviation (σ) of milk prices is $1.75. Let X be a random variable denoting milk price at super market. We plan to take a random sample of 36 supermarkets. Question 28^0 / 1^ pts What is the sampling distribution of x̄ when sample of size 36 is used? ou Answeredou Answered Is not normal because the sample size is too small Is normal due to the Central Limit Theorem orrect Answerorrect Answer Is not normal because the sample size is too large Is normal due to the Chebyshev’s Theorem
Refer Central Limit Theorem form Chap 7.(n>30) Question 29^0 / 1^ pts Suppose that we reduce the sample size from 36 to 16. The sampling distribution of x̄ will be normal only if ou Answeredou Answered X has a skewed distribution X has a uniform distribution orrect Answerorrect Answer X has a normal distribution X has a bi-modal distribution The sampling distribution will be normal only if X has a normal distribution when n Question 30^0 / 1^ pts What is the probability that a random sample of 36 supermarkets will provide an average milk price (x̄) that is more than $3.50? ou Answeredou Answered 45.5% orrect Answerorrect Answer 4.3%
