Math Is Fun Forum

Hi guys, I'm new to the forum and I just had a few question that I want to get help on. Some of the I've done and I want to know if the answer is corret, whereas for others, I want to know what to do. THankssss

1. An automobile club comes to the aid of stranded motorists who are its members. Over the long run, about 5%, which is the population proportion, of the members utilize this service in any 12-month period. A small town has 500 members of the club. One can consider these 500 members as a representative sample of the larger club membership. Let X and ˆp be respectively the number and proportion of the members in the town who will utilize the service in the coming 12-month period. [15]
(a) Name the distribution of X and give the values of any parameters involved.

X~Bin(500, 0.05)

(b) The distribution of X can be approximated by a continuous distribution. Name this
continuous distribution and give the values of any parameters involved.

Continuous random variable (Which continuous distribution, normal or t?)

(c) The towing service in the town has contracted with the club to come to the aid of
up to 35 members in the next 12-month period. What is the approximate probability
that the actual number of members requiring aid will exceed 35? (Use the continuous
distribution obtained in (b) and continuity correction to do the approximation.)

SD(X)=sqrt(np(1-p))=sqrt(500x0.05x0.95)=4.873

mu = 500 x 0.05
= 25

Z=(35.5-25)/4.873
= 2.1547

(d) Use Minitab to find the exact probability that more than 35 members in the town
will utilize the service in the next 12-month period. (Write down the answer and the
commands you used.)

(e) Calculate the mean and standard deviation of the sampling distribution of ˆp, namely,calculate E(ˆp) and sd(ˆp). Also name the (approximate) sampling distribution of ˆp.

(f) Suppose that 24 members in the town actually utilize the service in the next 12-months. Use this information to find the standard error of ˆp.

2. A survey of a class of 190 statistics students at a large university found that they averaged 7.1 hours of sleep the previous night. [10]
(a) Assume that the population average for adults is 6 hours and 57 minutes, or 6.95 hours of sleep (per night), with a (population) standard deviation of 2 hours. What is the (approximate) sampling distribution of the sample mean time of sleep for a random sample of 190 adults?

s.d.(x)= sigma/√n
= 2/√190
= 2/13.784
= 0.145

(b) How likely (i.e., what is the probability) that a sample mean of 7.1 hours of sleep or more would be observed from a random sample of 190 adults? Based on your answer, would the mean 7.1 hours of sleep obtained from the statistics students be a reasonable value to be expected for the sample mean of a random sample of 190 adults?

Pr(X≥7.1)
Z= (7.1-6.95)/2
=0.075
0.07 is 0.5279?
And 0.08 is 0.5319?

Which is about 52%/53%

It is expected because about 50% can be over an d50% can be under.

(c) Now focus on the population of university students in regard to sleep time (per night). Suppose its population mean sleep time is 6.95 hours, the same as that for adults in general population. But its population standard deviation of sleep time is unknown. Instead, the sample standard deviation of 1.95 hours is observed from the sample of 190 students.
i. Using this as an estimate of the student population standard deviation of sleep
time, write a standardized statistic for the sample mean sleep time of a random
sample of 190 students.
Z = ( x- mu)/(sigma/√n)
= (7.1-6.95)/1.95
= 0.0769

ii. What is the standard error of the sample mean sleep time of the 190 students?

s.e.= s/√n
= 1.95^2/√190
= 3.8025/13.784
= 0.276

iii. What is the name of the (approximate) sampling distribution for your standardized
statistic?

Standardizes z-statistic

iv. Use this distribution and Minitab to compute the probability that a sample mean of
7.1 hours of sleep or more would be observed from another sample of 190 university
students.

3. In a study done in Maryland, investigators surveyed individuals by telephone about how often they get tension headaches (Schwartz et al., 1998). One response variable measured was whether or not the respondent had experienced an episodic tension-type headache (ETTH) in the prior year. A headache pattern was called episodic if the headaches occurred less often than 15 times a month; otherwise, the headaches were called chronic. Of the 1600 women in the survey aged 18 to 29 years old, 653 said they had experienced episodic tension-type headaches in the last year. Of the 2122 women in the 30-to-39-year-old age group, the number of having experienced episodic headaches was 995. [15]

(a) Compute a 90% confidence interval, based on a normal distribution approximation, for the proportion of 18-to-29-year-old women in the population who experienced episodic headaches in the last year.

(b) Use the data given to test a hypothesis to check whether the population proportion
of 18-to-29-year-old women who experienced episodic headaches in the last year is less than 0.43. The test should be based on normal distribution and significance level 0.05.
You need to list in your work the null and alternative hypotheses, the test statistic, the p-value or rejection region, and a conclusion.

(c) Compute a 95% confidence interval for the difference between the proportions of episodic headaches for the two age groups in the population.

(d) Test a hypothesis at significance level  = 0.01 to check whether the difference between the two population proportions of episodic headaches is different from 0. You need to list in your work the null and alternative hypotheses, the test statistic, the p-value or rejection region, and a conclusion.

(e) Now suppose we want a 90% confidence interval for the population proportion of 30- to-39-year-old women who experienced episodic headaches in the last year to have a margin of error not more than 0.015. How many 30-to-39-year-old women should be included in the sample at least?