Question

To test Hy = 35 versus H, #35 a simple random sample of size 40 is obtained Complete parts achow Click the icon to view the table of critical t-values (a) Does the population have to be normally distributed to test this hypothesis by using distribution methods? Why? O A No-there are no constraints in order to perform a hypothesis test OB Yes-since the sample size is at not least 50, the underlying population does not need to be normally distributed OC. Yes--the population must be normally distributed in all cases order to perform a hypothesis test OD. No-since the sample size is at least 30, the underlying population does not need to be normally detributed (b)x= 387 and 5 =96, compute the test statistic - (Round to two decimal places as needed.) (c) Draw at distribution with the area that represents the P-value shaded. Choose the correct graph below ОА OB D. A A os (d) Approximate and interpret the P-value The probability of observing a statistic the one observed asing is true is in the range (e) If the researcher decides to test this hypothesis at the 0.01 level of significance will the researcher at the full hypothesis? Why? Because the P-value is than at the researcher wil the nul hypothesis (1) Construct a 99% confidence interval to test the hypothesis The lower bound is The upper bound is (Round to three decimal places as needed) Click to select your answers)

Answer 1

a) The population has to be normally distributed to test this hypothesis by using t distribution methods (as normal distribution of the population is the basic assumption for performing t test)

Since, the sample size is 40 (which is greater than 30, hence, a large sample size), so, by central Limit theorem, the underlying population can be assumed to be normally distributed.

Hence, option (D) us the correct option.

No, since, the sample size is atleast 30, the underlying population doesn't need to be normally distributed.

b) Null hypothesis (Ho) :

= 35

Alternative hypothesis (H1) : $\mu \neq 35$

Test statistic is given by -

CHO t s/n

where, is the sample mean = 38.7

T

s is the sample standard deviation = 9.6

n is the sample size = 40

is the specified value of population mean under the null hypothesis = 35

HO

Hence, the value of the test statistic will be -

38.7 – 35 t = 9.6/V40

= 2.43

Degrees of freedom = n - 1 = 40 - 1 = 39

c) P value = P(|t| > 2.43)

= Area under the t curve with 39 degrees of freedom on the left side of t = -2.43 and on the right side of t = 2.43

Hence, curve (c) will represent t distribution with area that represents P value shaded.

d) P value is the probability of observing the test statistic as extreme as the one observed, assuming null hypothesis (Ho) is true.

P value can be obtained using the formula =TDIST(2.43,39,2) in Excel and is equal to 0.0198

P value is in the range between 0.01 and 0.02

e) Because, the P value (0.0198) > $\alpha$ or level of significance (0.01), the researcher will not reject the null hypothesis.

f) 99% confidence interval is given by -

to.01/2,39 * n

where, $t_{0.01/2,39}$ is the critical value of t with 39 degrees of freedom for two tailed test at 0.01 level of significance = 2.71

(It can be obtained using the formula =TINV(0.01,39) in Excel)

Hence, confidence interval will be -

$[38.7 \pm 2.71 *9.6/\sqrt{40}]$

$=[38.7 \pm 4.11]$

= [ 34.587, 42.813]

Lower bound is 34.587

Upper bound is 42.813