Hello. Would you mind answering my questions? Thank you :) (Statistics) 1. If a

Question

Hello. Would you mind answering my questions? Thank you :)

(Statistics)

1. If a population distribution is very skewed (one tail is much longer than the other), then the t-test will not produce valid confidence intervals when the sample size is small (e.g. n < 20).

TRUE or FALSE ?

2. If a population distribution is very skewed, then we can obtain approximately valid confidence intervals by taking a large sample (e.g. n > 100).

  TRUE or FALSE?

3. You perform a t-test on some new data in R (R Studio), and obtain the following output:

        Two Sample t-test

data: sampA and sampB

t = 0.4084, df = 23, p-value = 0.6867

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-0.6306658 0.9409514

sample estimates:

   mean of x    mean of y

0.151746086 -0.003396702

Summarize the results of this analysis by describing

i) What hypothesis was tested?

ii) What kind of test was performed?

iii) What is the resulting p-value, and is it a one-sided or two-sided p-value?

iv) What is the confidence interval, and what parameter is it an interval for?

4. Suppose a random sample of size m = 20 female cats (X) has a sample mean weight of 10 pounds (lbs) and sample variance of 5 lbs2, and an independent sample of size n = 12 male cats (Y) has a sample mean weight of 9.7 lbs and a sample variance of 4 lbs2.

a) What is the pooled variance estimate sP2?

b) What is the t-statistic for testing the null hypothesis that the difference in population mean weights between male and female cats is 0?

c) How many degrees of freedom does the reference t-distribution have?

d) What critical value should you use to construct a 95% confidence interval for      X – Y?

e)Find a two-sided p-value for a test of H0: X – Y.

Thank you for your help!

Explanation / Answer

Q1)

Answer: True

Explanation: The normality assumption is not satisfying here.

Q2)

Answer: True

Explanation: As per central limit theorem in such cases it will follow approximately normal distribution.

Q3)

· The hypothesis tested was the true difference in means is not equal to 0.

· Two samples t-test for mean differences was performed.

· The 95% confidence interval is (-0.631, 0.941). The parameter is population mean differences.

Q4)

a) Pooled variance = [ (m-1)*sx^2 + (n-1)*sy^2 ] / ( m+n-2)

                            = (((20-1)*5) + ((12-1)*4) ) / (20+12-2)

                              = 4.6333

b) Test Statistics:

                      t = (x1 bar – x2 bar) / Ö [ sP^2 *((1/m)+(1/n)]

                         = (10-9.7)/Ö(4.6333 * ((1/20)+(1/12))

                         = 0.382

c) Df = m + n -2 = 20 + 12-2 = 30

d) 2.042 is the critical value for 30 degrees of freedom at 95% level of confidence.

e) Excel function of p-value =TDIST(0.382,30,2)

p-value = 0.7052

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