Question

The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey 8 employees. The number of sick days they took for the past year are as follows: 12; 6; 15; 5; 11; 10; 6; 8. Let X = the number of sick days they took for the past year. Should the personnel team believe that the mean number is about 10? Conduct a hypothesis test at the 5% level. Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) A.) State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom.) B.) What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.) C.) What is the p-value? (Round your answer to four decimal places.) D.) Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion. (i) Alpha (Enter an exact number as an integer, fraction, or decimal.) E.) Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to three decimal places.)

Answer 1

Given that,
population mean(u)=10
sample mean, x =9.125
standard deviation, s =3.4821
number (n)=8
null, Ho: μ=10
alternate, H1: μ!=10
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.365
since our test is two-tailed
reject Ho, if to < -2.365 OR if to > 2.365
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =9.125-10/(3.4821/sqrt(8))
to =-0.7107
| to | =0.7107
critical value
the value of |t α| with n-1 = 7 d.f is 2.365
we got |to| =0.7107 & | t α | =2.365
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 != -0.7107 ) = 0.5002
hence value of p0.05 < 0.5002,here we do not reject Ho
ANSWERS
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A.
assume that the underlying population is normally distributed
t test for single mean with unknown population standard deviation.
null, Ho: μ=10
alternate, H1: μ!=10
B.
test statistic: -0.7107
critical value: -2.365 , 2.365
decision: do not reject Ho
C.
p-value: 0.5002
D.
we do not have enough evidence to support the claim that the mean number is about 10.
E.
TRADITIONAL METHOD
given that,
sample mean, x =9.125
standard deviation, s =3.4821
sample size, n =8
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 3.4821/ sqrt ( 8) )
= 1.231
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 7 d.f is 2.365
margin of error = 2.365 * 1.231
= 2.912
III.
CI = x ± margin of error
confidence interval = [ 9.125 ± 2.912 ]
= [ 6.213 , 12.037 ]
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DIRECT METHOD
given that,
sample mean, x =9.125
standard deviation, s =3.4821
sample size, n =8
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 7 d.f is 2.365
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 = [ 9.125 ± t a/2 ( 3.4821/ Sqrt ( 8) ]
= [ 9.125-(2.365 * 1.231) , 9.125+(2.365 * 1.231) ]
= [ 6.213 , 12.037 ]
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interpretations:
1) we are 95% sure that the interval [ 6.213 , 12.037 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean