Question

In: Statistics and Probability

Math 120 Stats HW (sorry for posting multiples this is my last month of school and...

Math 120 Stats HW (sorry for posting multiples this is my last month of school and i cant afford another month of chegg..)

The following data represents at which babies crawl based on a survey of 12 moms. The data is normally distributed and s=10.698 weeks. Construct an interpreter a 95% confidence interval for the population standard deviation of the age and weeks at which babies first crawl.

55,32,44,34,38,26,48,36,59,27,39,29

A. If the repeated samples are taken, 95% of them will have the sample standard deviation between _ and _.

B. There is a 95% prob that the true population standard deviation is between _ and _

C. There is 95% confidence that the population SD is between _ and _

and

The following data represents the pH of rain for a sample of 12 rain dates. A normal probability plot suggests the data could come from a population that is normall distributed. boxplot says there are no outliers. 5.20, 5.72, 4.38, 4.80, 5.02, 4.58, 4.74, 5.19, 5.34, 4.76, 4.56, 5.68

B. Construct and interpret a 95% confidence interval for the mean pH of rainwater. Select the correct choice below and complete the boxes below.

a. there is 95% confidence that the pop mean of ph of rain water is between _ and _.

b. if repeated samples are taken, 95% of them will have a sample ph of rain water between _ and _

c. there js a 95% prob that the true mean of rain water is between _ and _
----------------------------------------------------
section C
construct and interpret a 99% confidence interval for the mean ph pf rainwater.

a. there is 99% confidence that the population mean ph of rain water is between _ and _

b. theres a 99% chance that the true mean of ph water is _ and _

c. if repeated samples are taken, 99% of them will have a sample ph of water between _ and _

Solutions

Expert Solution

Solution-A:

For the given sample ,

n=12

df=n-1=12-1=11

sample standaed dev=s=10.69799

alpha/2=0.05/2=0.025

1-alpha/2=1-0.025=0.975

chi sq alpha/2==CHISQ.INV.RT(0.025,11)=21.92004926

chi sq 1-alpha/2==CHISQ.INV.RT(0.975,11)=3.815748252

95% confidence interval for sigma

sqrt(n-1)*s^2/chisq(1-alpha/2)<sigma<sqrt(n-1)*s^2/chisq(alpha/2)

sqrt((12-1)*10.69799^2/21.92004926)<sigma<sqrt((12-1)*10.69799^2/3.815748252)

7.578404<sigma<  18.16388

7.578404<sigma<  18.16388

There is 95% confidence that the population SD is between 7.578 and 18.164

Solution-b:
Rcode:

ph <- c(5.20, 5.72, 4.38, 4.80, 5.02, 4.58, 4.74, 5.19, 5.34, 4.76, 4.56, 5.68)
t.test(ph,conf.level = 0.95)

output:


data: ph
t = 39.693, df = 11, p-value = 3.147e-13
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
4.720388 5.274612
sample estimates:
mean of x
4.9975

ANSWER;

95 percent confidence interval:
4.720388 5.274612

there is 95% confidence that the pop mean of ph of rain water is between 4.720388 and 5.274612

Solution-c:

ph <- c(5.20, 5.72, 4.38, 4.80, 5.02, 4.58, 4.74, 5.19, 5.34, 4.76, 4.56, 5.68)
t.test(ph,conf.level = 0.99)
   One Sample t-test

data: ph
t = 39.693, df = 11, p-value = 3.147e-13
alternative hypothesis: true mean is not equal to 0
99 percent confidence interval:
4.606468 5.388532
sample estimates:
mean of x
4.9975

ANSWER::

99 percent confidence interval:
4.606468 and 5.388532

a. there is 99% confidence that the population mean ph of rain water is between  4.606468 and 5.388532


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