Question

To test Ho: = 50 versus H=50, a simple random sample of size n = 40 is obtained. Complete parts (a) through below 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 t-distribution methods? Why? O A. No-there are no constraints in order to perform a hypothesis test. O B. No-since the sample size is at least 30, the underlying population does not need to be normally distributed. OC. Yes since the sample size is at not least 50, the underlying population does not need to be normally distributed. OD Yes-the population must be normally distributed in all cases in order to perform a hypothesis test. (b) if x = 53.5 and s= 3.8. compute the test statistic t = 2.57 (Round to two decimal places as needed.) (c) Draw a t-distribution with the area that represents the P-value shaded. Choose the correct graph below. OA ов. Ос. OD ^ ^ (d) Approximate and interpret the P-value. The probability of observing a population statistic as extreme or less extreme than the one observed, assuming His true, is in the range 0.02<P-value < 0.04 (e) If the researcher decides to test this hypothesis at the a= 0.01 level of significance, will the researcher reject the null hypothesis? Why? Because the P-value is less than , the researcher will fail to reject the null hypothesis. (f) Construct a 99% confidence interval to test the hypothesis. The lower bound is? The upper bound is? (Round to three decimal places as needed.) reject the null hypothesis. Because the value lies within the confidence interval, we (Type an integer or decimal. Do not round.)

Answer 1

Given that,
population mean(u)=50
sample mean, x =53.5
standard deviation, s =8.6
number (n)=40
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =2.708
since our test is two-tailed
reject Ho, if to < -2.708 OR if to > 2.708
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =53.5-50/(8.6/sqrt(40))
to =2.574
| to | =2.574
critical value
the value of |t α| with n-1 = 39 d.f is 2.708
we got |to| =2.574 & | t α | =2.708
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.5739 ) = 0.014
hence value of p0.01 < 0.014,here we do not reject Ho
ANSWERS
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a.
option:D
population is normally distributed
null, Ho: μ=50
alternate, H1: μ!=50
b.
test statistic: 2.57
c.
option:C
as diagram shows
critical value: -2.708 , 2.708
decision: do not reject Ho
d.
the probability of observing a population statistic as extreme as the one observed ,assuming H1 is true the range p value is 0.01 to 0.04
e.
p-value: 0.014
p value is greater than alpha value.
we do not have enough evidence to support the claim that population mean is equal to 50.
f.
TRADITIONAL METHOD
given that,
sample mean, x =53.5
standard deviation, s =8.6
sample size, n =40
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 8.6/ sqrt ( 40) )
= 1.36
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 39 d.f is 2.708
margin of error = 2.708 * 1.36
= 3.682
III.
CI = x ± margin of error
confidence interval = [ 53.5 ± 3.682 ]
= [ 49.818 , 57.182 ]
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DIRECT METHOD
given that,
sample mean, x =53.5
standard deviation, s =8.6
sample size, n =40
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 39 d.f is 2.708
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 53.5 ± t a/2 ( 8.6/ Sqrt ( 40) ]
= [ 53.5-(2.708 * 1.36) , 53.5+(2.708 * 1.36) ]
= [ 49.818 , 57.182 ]
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interpretations:
1) we are 99% sure that the interval [ 49.818 , 57.182 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
99% sure that the interval [ 49.818 , 57.182 ] , population mean is lies within the interval