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Homework answers / question archive / The amount of time spent by American adults playing sports per day is normally distributed with a mean of 4 hours and standard deviation of 1
The amount of time spent by American adults playing sports per day is normally distributed with a mean of 4 hours and standard deviation of 1.25 hours.
17. Find the probability that a randomly selected American adult plays sports for more than 5 hours per day
18. Find the probability that if four American adults are randomly selected, their average number of hours spent playing sports is more than 5 hours per day.
19. Find the probability that if four American adults are randomly selected, all four play sports for more than 5 hours per day.
20. A random sample of 10 waitresses in Iowa City, Iowa revealed the following hourly earnings
(including tips):
19 18 15 16 18 17 16 18 20 14
If the hourly earnings are normally distributed with a standard deviation of $4.5, estimate with 95%
confidence the mean hourly earnings for all waitresses in Iowa City.
21. A statistician wants to estimate the mean weekly family expenditure on clothes. He believes that the standard deviation of the weekly expenditure is $125. Determine with 99% confidence the number of families that must be sampled to estimate the mean weekly family expenditure on clothes to within
$15.
22. The temperature readings for 20 winter days in Grand Rapids, Michigan are normally distributed with a mean 5.5 degrees and a standard deviation of 1.5. Determine the 90% confidence interval estimate for the winter mean temperature.
24. A social scientist claims that the average adult watches less than 26 hours of television per week. He collects data on 25 individuals' television viewing habits and finds that the mean number of hours that the 25 people spent watching television was 22.4 hours. If the population standard deviation is known to be eight hours, can we conclude at the 1% significance level that he is right?
Questions 25 through 27 are based on the following information:
During the last energy crisis, a government official claimed that the average car owner refilled the tank when there was more than 3 gallons left. To check the claim, 10 cars were surveyed as they entered a gas station. The amount of gas was measured and recorded as shown below
3 5 3 2 3 3 2 6 4 1
Assume that the amount of gas remaining in tanks is normally distributed with a standard deviation of 1 gallon.
25. Can we conclude at the 10% significance level that the official was correct?
26. Calculate the p-value
Questions 27 through 28 are based on the following information:
A simple random sample of 100 observations was drawn from a normal population. The mean and standard deviation of the sample were 120 and 25, respectively.
27. Test the following hypotheses, with α = 0.10:
H0: μ = 125 H1: μ ≠ 125
28. Estimate the population mean with 90% confidence.
29. Given that a random sample of size 120 revealed that the number of successes was 57, test the hypotheses below at the 5% significance level:
H0: p = 0.40 H1: p ≠ 0.40
30. A union composed of several thousand employees is preparing to vote on a new contract. A random sample of 500 employees yielded 320 who planned to vote yes.
a. Can we infer at the 5% significance level that the new contract will receive more than 60% yes votes?
b. Compute the p-value for the test in part (a)
17) p(x > 5) =(x - miu)/sigma
p(x > 5) =(5 -4)/1.25 =0.8
z score of 0.8 is 0.78814
p(x > 5) =1 -0.78814 =0.21186
18) p(x > 5) =(x - miu)/[sigma/sqrt(n)]
p(x > 5) =(5-4)/[1.25/sqrt(4)] = 1/0.625 =1.6
p value of 1.6 from t table = 0.104
p(x > 5) =0.104
19) 0.104 x 4 =0.416
20) 95% CI = x bar + or - t [s/sqrt(n)]
x bar =171/10 =17.1
t0.05, 9 =2.262
95% CI = 17.1 + or - 2.262[4.5/sqrt(10)]
95% CI =17.1 + or - 3.219
95% CI =(13.88, 20.32)
21) n =[(z x sigma)/e)]2
n =[(2.58 x 125)/15]2
n =462
22) 90% CI = x bar + or - t [s/sqrt(n)]
t0.10, 19 =1.729
90% CI =5.5 + or - 1.729[1.5/sqrt(20)
90% CI =5.5 + or - 0.58
90% CI =(4.92, 6.08)
24) H0: miu > or =26 hours
H1: miu < 26 hours
alpha =0.01
t =(x bar - miu)/[s/sqrt(n)]
t=(22.4 -26)/[8/sqrt(25)]
t =-3.6/1.6 =-2.25
t0.01, 24 =-2.492
Decision rule; Reject H0 if t < -2.492
Therefore we fail to reject H0 because -2.25 > -2.492
Hence we have no statistically significant evidence at alpha =0.01 to show that the average adult watches less than 26 hours
of television per week. The claim is rejected.
25) H0: miu < or =3 gallons
H1: miu > 3 gallons
alpha =0.10
x bar =32/10 =3.2
t=(x bar - miu)/[s/sqrt(n)]
t =(3.2 -3)/[1/sqrt(10)]
t =0.2/0.316 =0.632
t0.10, 9 =1.383
Decision rule; Reject H0 if t > 1.383
Therefore we fail to reject H0 because 0.632 < 1.383
Hence we have no statistically significance evidence at alpha =0.10 to show that the average car owner refilled the tank when there was more than 3 gallons left.
26) p-value of 0.632 from t table is 0.271553
27) H0: miu =125
H1: miu is not = 125
alpha =0.10
z =(x bar - miu)/[s/sqrt(n)]
z =(120 -125)/[25/sqrt(100)]
z =-5/2.5 =-2
Decision rule; Reject H0 if z < or =-1.645
Therefore we reject H0 because -2 < -1.645
Hence we have statistically significant evidence at alpha=0.10 to show that miu is not = 125
28) 90% CI =x bar + or - z [s/sqrt(n)]
90% CI =120 + or - 1.645[25/sqrt(100)]
90% CI =120+ 4.1125
90% CI =(115.888, 124.112)
29) H0: p =0.40
H1: p is not =0.40
p =57/120 =0.475
z =(p -p0)/sqrt[p0 (1-p0)/n]
z =(0.475 -0.4)/sqrt[0.4(0.6)/120]
z =0.075/0.0447 =1.677
Decision rule; Reject H0 if z > 1.96
Therefore we fail reject H0 because 1.677 < 1.96
Hence we have no statistically significant evidence at alpha =0.05 to show that proportion is not = 0.40
30) a) H0: p < or =0.60
H1: p > 0.60
alpha =0.05
p =320/500 =0.64
z =(p - p0)/sqrt[p0(1 -p0)/n]
z =(0.64 -0.60)/sqrt[0.6(0.4)/500]
z =0.04/0.022 =1.826
Decision rule; Reject H0 if z > 1.645
Therefore we reject H0 because 1.826 > 1.645
Hence we have statistically significant evidence at alpha =0.05 to show that the new contract will receive more than 60%
yes votes.
b) p value of 1.826 fro z table is 0.033925
