Question

THIS IS ONE QUESTION . THANKS

1) A random sample of 100 observations from a population with standard deviation 11.99 yielded a sample mean of 92

1. Given that the null hypothesis is ?=90 and the alternative hypothesis is ?>90 using ?=.05 find the following:

(a) Test statistic =

(b)  P - value:

B) Given that the null hypothesis is ?=90 and the alternative hypothesis is ?≠90 using ?=.05 find the following:

(a) Test statistic =

(b)  P - value:

2) A random sample of 10 observations was drawn from a large normally distributed population. The data is below.

23,19, 24, 17, 19, 23, 22, 22, 17, 16,

Test to determine if we can infer at the 8% significance level that the population mean is not equal to 20, filling in the requested information below.

A. The value of the standardized test statistic:

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Note: For the next part, your answer should use interval notation. An answer of the form (−∞,?) is expressed (-infty, a), an answer of the form (?,∞)is expressed (b, infty), and an answer of the form (−∞,?)∪(?,∞) is expressed (-infty, a)U(b, infty).

B. The rejection region for the standardized test statistic:

C. The p-value is

Answer 1

1.

1. The null hypothesis H0:µ=90 vs the alternative hypothesis H1: µ>90

Given the sample size n=100

the population standard deviation =ơ=11.99

Sample mean=X̄=92

(a) For the above one-sample mean test for known population standard deviation, the test statistic is:

Z={√n*( X̄-µ0})/ơ

where µ0 is the hypothesized mean =90

So, test statistic:

Z={√n*( X̄-µ0})/ơ

={√100 *(92-90)}/11.99

=20/11.99

=1.67

Ans:1.67

(b)

The above test statistic follows Z distribution

For the one-sided test (Since alternative is of ">" type), The p-value for Z=1.67 is given by

p=P( Z>1.67)=0.047 {From normal table}

Ans :0.047

B. The null hypothesis H0:µ=90 vs the alternative hypothesis H1: µ≠90

(a) For this one-sample mean test, the conditions are same as in the 1st part, only the alternative hypothesis is two-tailed (i.e the altenative is of "not equal to" type)

The test statistic is

Z={√n*( X̄-µ0})/ơ

that is same as in part 1

So test statistic =1.67 ( From part 1)

Ans:1.67

(b)

The above test statistic follows Z distribution

For the two-sided test, The p-value for Z=1.67 is given by

p=P(|Z| > 1.67) =P(Z < -1.67) +P( Z> 1.67) =1-P(Z<1.67)+P(Z>1.67) =1-0.953+0.047 =0.094 [From normal table]

Ans :0.094

* Note for part 1, null hypothesis will be rejected as p-value < alpha=0.05. However, in part 2 , we will fail to reject null hypothesis if p-value > alpha=0.05

2. The null hypothesis H0:µ=20 vs the alternative hypothesis H1: µ≠20

alpha=0.08

Here the population standard deviation is unknown.

A. The test statistic is given by:

t=

the sample size n=10

Sample mean=X̄=1/n*∑xi =1/10*(23+19+..+17.+16)=1/10*202=20.2

Sample variance=s=√{1/(n-1)*∑(xi-X̄)^2} =√{1/9*(23-20.2)^2+(19-20.2)^2+....+(16-20.2)^2} =2.9364

µ0 is the hypothesized mean =20

So the test statistic is

=√n*( X̄-µ0})/s

=√10*(20.2-20)/2.9364

=0.2154

Ans :0.2154

B. The above test statistic follows a t distribution with (n-1) degrees of freedom.

The rejection region for the two-tailed test is

|t| > 0.2154

C.So the p-value for t=0.2154 with degrees of freedom=9 is

P(|t| > 0.2154) =0.834 [ From calculator]

Ans: 0.834