Humans are known to have a mean gestation period of 280 days (from last menstrua

Question

Humans are known to have a mean gestation period of 280 days (from last menstruation) with a standard deviation of about 9 days. A hospital wondered whether there was any evidence that their patients were at risk for giving birth prematurely. In a random sample of 70 women from this hospital, the average gestation time was 274.3 days. Can you conclude that the sample data provides enough evidence that the average gestation time of all the patients at this hospital was less than 280 days at a significant level of 0.05? Use the 8-step critical region approach. Can you use the P-value to conclude that the sample data provides enough evidence that the average gestation time of all the patients at this hospital was less than 280 days at a significant level of 0.05? Why or why not? Can you conclude that the sample data provides enough evidence that the average gestation time of all the patients at this hospital was less than 280 days at a significant level of 0.05 using an appropriate confidence interval? State you reason. a. b. c.

Explanation / Answer

a & b.)
Given that,
population mean(u)=280
standard deviation, =9
sample mean, x =274.3
number (n)=70
null, Ho: =280
alternate, H1: <280
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 274.3-280/(9/sqrt(70)
zo = -5.3
| zo | = 5.3
critical value
the value of |z | at los 5% is 1.645
we got |zo| =5.3 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -5.3 ) = 0
hence value of p0.05 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: =280
alternate, H1: <280
test statistic: -5.3
critical value: -1.645
decision: reject Ho
p-value: 0

we have evidence that the average gestation time of all the patients is
less than 280

c.
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 274.3 ± Z a/2 ( 9/ Sqrt ( 70) ) ]
= [ 274.3 - 1.96 * (1.076) , 274.3 + 1.96 * (1.076) ]
= [ 272.192,276.408 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [272.192 , 276.408 ] 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

using confidence we have evidence that the average gestation time of all the patients is
less than 280

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.