Estimation and Test of Hypothesis

Solution Preview

---

Please see attached file
1 When the means of two unrelated samples are used to compare two populations, we are dealing with two dependent means ___ T/F

F
We are dealing with independent means as the samples are unrelated.

2 If s is unknown when completing a hypothesis test about the population mean, then the best estimate for the unknown standard deviation is s. _____ T/F

T

3 The number of degrees of freedom for the critical value of t is equal to the smaller of n1 -1 or n2 -1 when making inferences about the difference between two independent means for the case when the degrees of freedom are estimated. ___ T/F

F
degrees of freedom = n1+n2-2

4 Pretest versus post-test (before versus after) studies are usually independent samples. ____ T/F

F
These are dependent samples

5 The chi-square distribution is used for inferences about the population mean M when the standard deviation s is unknown. ____ T/F

F
The t distribution is used for inferences about the population mean M when the standard deviation s is unknown and sample size is small.

6 The t-distribution is used for all inferences about a population's variance. ___ T/F

F
The chi square distribution is used for inferences about a population's variance

7 The F distribution is a symmetric distribution. ____ T/F

F
The F distribution is an asymmetric distribution.

8 Which of the following would be the correct hypotheses for testing the claim that the mean waiting time to be served at a large post office is at least 6.5 minutes?
A. B. H0: M = 6.5 and Ha: M not equal to 6.5
B. H0: M <= 6.5 vs. Ha: M > 6.5
C. H0: M >= 6.5 and Ha: M < 6.5
D. H0: M > 6.5 vs. Ha: M >= 6.5

Answer: B. H0: M <= 6.5 vs. Ha: M > 6.5

This is a right tailed test as we are testing the hypothesis that the mean waiting time is more than 6.5 minutes
The rejection region is the right tail.

9 Which of the following would be the correct hypotheses for testing the claim that the proportion of male golfers (M) at a particular college is no more than the proportion of female golfers (F) at the college?
A. H0: pM >= pF
B. H0: pM <= pF
C. H0: pM > pF
D. H0: pM < pF

Answer: A. H0: pM >= pF

The research (alternative) hypothesis is that pM<pF
The null hypothesis H0 is therefore pM>=pF

10 The mean age of 25 randomly selected college seniors was found to be 23.5 years, and the standard deviation of all college seniors was 1.3 years. Which of the following is the correct symbol for 23.5 years?
A. M
B. s
C. s
D.

Answer: D.

23.5 is the

Please show all calculations on every problem.

11 A machine produces 3-inch nails. A sample of 12 nails was selected and the lengths determined. The results are as follows:
2.89 2.95 3 3.05 2.99 2.96 3.1 3.06 3 3.12 3 2.95

Use these results to test H0: M = 3 and Ha: M not equal to 3 at alpha (a) = 0.05. Give the critical region, the computed test statistic, and the conclusion.

We first caculate the sample mean and sample standard deviation
Mean and standard deviation

X= X 2 =
2.89 8.35
2.95 8.70
3.00 9.00
3.05 9.30
2.99 8.94
2.96 8.76
3.10 9.61
3.06 9.36
3.00 9.00
3.12 9.73
3.00 9.00
2.95 8.70

Total= 36.07 108.4693
n=no of observations= 12
Mean= 3.0058 =36.07/12

variance={summation of X 2 - (summation of X) 2 /n}/(n-1)= 0.004445 =(108.4693-36.07^2/12)/(12-1)
standard deviation =square root of Variance= 0.0667 =square root of 0.004445

Data
Hypthesized mean= 3 inches
Sample Standard deviation= 0.0667 inches (calculated above)
Sample mean= 3.0058 inches (calculated above)
Sample size= 12
Significance level= 0.05 0r 5%
Population standard deviation (Known, Not Known)= Not Known

1) Hypothesis
Null Hypothesis: Ho: M = 3 (:Mean is 3 inches)
Alternative Hypothesis: H1: M not equal to 3 :( Mean is not equal to 3 inches )
Significance level=alpha (a) = 0.05 or 5%
No of tails= 2 (Both tails )
This is a 2 tailed (Both tails ) test because we are testing that M not equal to 3

2) Decision rule
sample size=n= 12
Since sample size 12 is less than 30 and poulation standard deviation is 'Not Known' use t distribution
Thus we use t distribution
t at the 0.05 level of significance and 11 degrees of freedom (=n-1) and 2 tailed test= 2.201
t critical = + or - 2.201

if sample statistic is <-2.201 or > 2.201 Reject Null Hypothesis, else Accept Null Hypothesis
Alternatively,
if p value is less than the significance level (= 0.05 ) Reject Null Hypothesis, else Accept Null Hypothesis

3) Calculation of sample statistics

Hypothesized Mean=M = 3 inches
Standard deviation =s= 0.0667 inches
sample size=n= 12
sx=standard error of mean=s/square root of n= 0.0193 = ( 0.0667 /square root of 12)
sample mean= 3.0058 inches

t=(sample mean-M )/sx= 0.3005 =(3.0058-3)/0.0193

4) Compare sample statistic with critical value

test statistic= 0.3005
t critical = + or - 2.201

therefore, sample statistic is within acceptance region

5) Decision

Accept Null Hypothesis (:Mean is 3 inches)

Alternatively, calculate p value
Prob-value corresponding to t = 0.3005 is 76.94% ...