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PIT_PIT [208]
3 years ago
10

A random sample of size 25 is taken from a nor- mal population having a mean of 80 and a standard deviation of 5. A second rando

m sample of size 36 is taken from a different normal population having a mean of 75 and a standard deviation of 3. Find the probability that the sample mean computed from the 25 measurements will exceed the sample mean com- puted from the 36 measurements by at least 3.4 but less than 5.9. Assume the difference of the means to be measured to the nearest tenth.
Mathematics
2 answers:
BabaBlast [244]3 years ago
7 0

Answer:

4

Step-by-step explanation:

Gekata [30.6K]3 years ago
3 0

Answer:  .7073

Step-by-step explanation:

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A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute
Katyanochek1 [597]

Answer:

Step-by-step explanation:

Given that:

sample size n = 36

standard deviation = 10.1

level of significance ∝ = 0.10

The null hypothesis and the alternative hypothesis can be computed as follows:

H_0 : \sigma = 10

H_1 : \sigma \neq 10

The test statistics can be computed as follows:

X^2 = \dfrac{(n -1)s^2 }{\sigma ^2}

X^2 = \dfrac{(36 -1)10.1^2 }{10^2}

X^2 = \dfrac{(35)102.01 }{100}

X^2 = \dfrac{3570.35 }{100}

X^2 =35.704

degree of freedom = n - 1 = 36 - 1 = 35

Since this test is two tailed .

The P -value can be determined by using the EXCEL FUNCTION ( = 2 × CHIDIST(35.7035, 35)

P - value = 2 × 0.435163515

P - value = 0.8703      ( to four decimal places)

Decision Rule : To reject the null hypothesis if  P - value is less than the 0.10

Conclusion: We fail to reject null hypothesis ( accept null hypothesis) since p-value is greater than 0.10 and we conclude that there is sufficient claim that the normal range of pulse rates of adults given as 60 to 100 beats per minute resulted to a standard deviation of 10 beats per minute.

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