Section 9.3: Tests on the Mean of a Normal Distribution Variance Unknown 1. Huma

Question

Section 9.3: Tests on the Mean of a Normal Distribution Variance Unknown 1. Humans are known to have a mean gestation period of 280 days. A hospital wondered whether there was any evidence that their patients were at risk for giving birth prematurely. In a random sample of 40 women, the average gestation time was 277.3 days with a standard deviation of 8.9 days. Use -001. a) State the hypotheses. b) Check the necessary assumptions. c) Calculate the test statistic. d) Find the p-value. e) Make a decision. f State the conclusion. g) Are the results statistically significant, at the 5% level? At the 1% level? h) Construct an appropriate confidence interval (CI) to test the hypotheses. State the decision based on the Cl. Does your decision match the decision in (e)?

Explanation / Answer

9.3.

Given that,
population mean(u)=280
standard deviation, =8.9
sample mean, x =277.3
number (n)=40
null, Ho: =280
alternate, H1: <280
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 277.3-280/(8.9/sqrt(40)
zo = -1.9187
| zo | = 1.9187
critical value
the value of |z | at los 1% is 2.326
we got |zo| =1.9187 & | z | = 2.326
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : left tail - ha : ( p < -1.9187 ) = 0.0275
hence value of p0.01 < 0.0275, here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: =280
alternate, H1: <280
b.
test statistic: -1.9187
c.
critical value: -2.326
d.
p-value: 0.0275
e.
decision: do not reject Ho
f.
we do not have enough evidence to support that were patients at risk for giving birth prematurely.
g.
results are significant at 1% level only
i.
TRADITIONAL METHOD
given that,
standard deviation, =8.9
sample mean, x =277.3
population size (n)=40
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 8.9/ sqrt ( 40) )
= 1.407
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
value of z table is 2.326
margin of error = 2.326 * 1.407
= 3.273
III.
CI = x ± margin of error
confidence interval = [ 277.3 ± 3.273 ]
= [ 274.027,280.573 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =8.9
sample mean, x =277.3
population size (n)=40
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
value of z table is 2.326
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 277.3 ± Z a/2 ( 8.9/ Sqrt ( 40) ) ]
= [ 277.3 - 2.326 * (1.407) , 277.3 + 2.326 * (1.407) ]
= [ 274.027,280.573 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [274.027 , 280.573 ] 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
[ANSWERS]
best point of estimate = mean = 277.3
standard error =1.407
z table value = 2.326
margin of error = 3.273
confidence interval = [ 274.027 , 280.573 ]

yes, decision is matching in (e)

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