In: Statistics and Probability
1) The population standard deviation for waiting times to be seated at a restaurant is known to be 10 minutes. An expensive restaurant claims that the average waiting time for dinner is approximately 1 hour, but we suspect that this claim is inflated to make the restaurant appear more exclusive and successful. A random sample of 30 customers yielded a sample average waiting time of 50 minutes.
2) A sample of 800 items produced on a new machine showed that 48 of them are defective. The factory will get rid the machine if the data indicates that the proportion of defective items is significantly more than 5%. At a significance level of 10% does the factory get rid of the machine or not?
3) A psychologist claims that the mean age at which children start walking is 12.5 months. The following data give the age at which 18 randomly selected children started walking.
15 11 13 14 15 12 15 10 16
17 14 16 13 15 15 14 11 13
4) According to a study, 107 of 507 female college students were on a diet at the time of the study.
a) Construct a 99% confidence interval for the true proportion of all female students who were on a diet at the time of this study. Insert the results from StatCrunch here.
b) Interpret this interval.
c) Is it reasonable to think that only 17% of college women are on a diet? Why or why not?
1)
The original data (individual waiting times) is not normally distributed. but central limit theorem allows us to do the calculations in part (a)
Ho : µ ≥ 60
Ha : µ < 60
(Left tail test)
population std dev , σ =
10.0000
Sample Size , n = 30
Sample Mean, x̅ = 50.0000
' ' '
Standard Error , SE = σ/√n = 10.0000 / √
30 = 1.8257
Z-test statistic= (x̅ - µ )/SE = (
50.000 - 60 ) /
1.8257 = -5.477
critical z value, z* =
-2.326 [Excel formula =NORMSINV(α/no. of tails)
]
p-Value = 0.0000 [ Excel
formula =NORMSDIST(z) ]
Decision: p-value<α, Reject null hypothesis
conclusion: there is enough evidence to say that the restaurant’s
claim is too high
2)
Ho : p = 0.05
H1 : p > 0.05
(Right tail test)
Level of Significance, α =
0.10
Number of Items of Interest, x =
48
Sample Size, n = 800
Sample Proportion , p̂ = x/n =
0.0600
Standard Error , SE = √( p(1-p)/n ) =
0.0077
Z Test Statistic = ( p̂-p)/SE = ( 0.0600
- 0.05 ) / 0.0077
= 1.2978
critical z value =
1.282 [Excel function =NORMSINV(α)
p-Value = 0.0972 [Excel
function =NORMSDIST(-z)
Decision: p-value<α , reject null hypothesis
There is enough evidence to conclude that the factory should get
rid of the machine