Question

1. A national newspaper reported that the state with the longest mean life span is Hawaii, where the population mean life span is 77 years. A random sample of 20 obituary notices in the Honolulu Advertizer gave the following information about life span (in years) of Honolulu residents.

72	68	81	93	56	19	78	94	83	84
77	69	85	97	75	71	86	47	66	27

(i) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.)

x	=	yr
s	=	yr

(ii) Assuming that life span in Honolulu is approximately normally distributed, does this information indicate that the population mean life span for Honolulu residents is less than 77 years? Use a 5% level of significance.

(a) What is the level of significance?
  

State the null and alternate hypotheses.

H₀: μ < 77 yr; H₁:  μ = 77 y

rH₀: μ = 77 yr; H₁:  μ > 77 yr    

H₀: μ = 77 yr; H₁:  μ ≠ 77 yr

H₀: μ = 77 yr; H₁:  μ < 77 yr

H₀: μ > 77 yr; H₁:  μ = 77 yr

(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

The Student's t, since we assume that x has a normal distribution and σ is unknown.The Student's t, since we assume that x has a normal distribution and σ is known.     The standard normal, since we assume that x has a normal distribution and σ is known.The standard normal, since we assume that x has a normal distribution and σ is unknown.

What is the value of the sample test statistic? (Round your answer to three decimal places.)
  

(c) Find the P-value. (Round your answer to four decimal places.)

2. The Student's t distribution table gives critical values for the Student's t distribution. Use an appropriate d.f. as the row header. For a right-tailed test, the column header is the value of α found in the one-tail area row. For a left-tailed test, the column header is the value of α found in the one-tail area row, but you must change the sign of the critical value t to −t. For a two-tailed test, the column header is the value of α from the two-tail area row. The critical values are the ±tvalues shown.

Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is μ = 7.4†. A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 36 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.0 with sample standard deviation s = 1.7. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood. Solve the problem using the critical region method of testing (i.e., traditional method). (Round your answers to three decimal places.)

test statistic	=
critical value	= ±

State your conclusion ion the context of the application.

Fail to reject the null hypothesis, there is sufficient evidence that the drug has changed the mean pH level.

Fail to reject the null hypothesis, there is insufficient evidence that the drug has changed the mean pH level.     

Reject the null hypothesis, there is insufficient evidence that the drug has changed the mean pH level.

Reject the null hypothesis, there is sufficient evidence that the drug has changed the mean pH level.

Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same?

We reject the null hypothesis using the traditional method, but fail to reject using the P-value method.

The conclusions obtained by using both methods are the same.    

We reject the null hypothesis using the P-value method, but fail to reject using the traditional method.

Answer 1

1.

Given that,
population mean(u)=77
sample mean, x =71.4
standard deviation, s =20.6484
number (n)=20
null, Ho: μ=77
alternate, H1: μ<77
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.729
since our test is left-tailed
reject Ho, if to < -1.729
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =71.4-77/(20.6484/sqrt(20))
to =-1.213
| to | =1.213
critical value
the value of |t α| with n-1 = 19 d.f is 1.729
we got |to| =1.213 & | t α | =1.729
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.2129 ) = 0.12002
hence value of p0.05 < 0.12002,here we do not reject Ho
ANSWERS
---------------
i.
sample mean, x =71.4
standard deviation, s =20.6484
number (n)=20
ii.
standard normal distribution
null, Ho: μ=77
alternate, H1: μ<77
b.
The Student's t, since we assume that x has a normal distribution and σ is unknown
c.
test statistic: -1.213
critical value: -1.729
decision: do not reject Ho
p-value: 0.12002
we do not have enough evidence to support the claim that the population mean life span for Honolulu residents is less than 77 years
2.
Given that,
population mean(u)=7.4
sample mean, x =8
standard deviation, s =1.7
number (n)=36
null, Ho: μ=7.4
alternate, H1: μ!=7.4
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.03
since our test is two-tailed
reject Ho, if to < -2.03 OR if to > 2.03
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =8-7.4/(1.7/sqrt(36))
to =2.118
| to | =2.118
critical value
the value of |t α| with n-1 = 35 d.f is 2.03
we got |to| =2.118 & | t α | =2.03
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.1176 ) = 0.0414
hence value of p0.05 > 0.0414,here we reject Ho
ANSWERS
---------------
null, Ho: μ=7.4
alternate, H1: μ!=7.4
test statistic: 2.118
critical value: -2.03 , 2.03
decision: reject Ho
p-value: 0.0414
we have enough evidence to support the claim that the drug has changed (either way) the mean pH level of the blood.
Compare your conclusion with the conclusion obtained by using the P-value method.
The conclusions obtained by using both methods are the same.