Let xi X17 be a sample of n-17 observations from normal deviation is s. We wish

Question

Let xi X17 be a sample of n-17 observations from normal deviation is s. We wish to test the null hypothesis Ho:p 15 against the alternative using R code. a)if xbar - 18.25 and s-2.31 then what is the value of T? b) T has a t distribution. How many degrees of freedom does this distribution have? c) Ef we test at the 1% level then what is the etical value(tstar)? d) What is the p-value based upon our sample? e) Do we reject the nut hypothesis at the 1% leven(Y/N)? a normal population with unknown mean ? and unknown standard deviation a . Suppose the sample mean is x and the sample standard Ma 15. Our test statistic wil be T 17 Our sample yleldsx- 16.75and s -2.31. Answer the following O Cannot Determine O Alternative O 2.31/sgrt(17) O Yes ? If we repeat . 1% level test 700 times then about how many tim. do we expect to nia typem eroa

Explanation / Answer

a)

t = (xbar - 15)/ (s /sqrt(n))

xbar = 18.25

s= 2.31

n=17

inserting the values of xbar and s,

t = 5.8009

b)

the degrees of freedom is n-1

df = n - 1 = 17 - 1 = 16

c)

alpha = 0.01

t-stat at 16 degrees of freedom at 0.01 alpha for a right tailed test can be calculated in excel as

=+T.INV(0.01,16)

-2.567

ignoring the sign it is 2.5.67

d)

p-value is t-stat of 16 degrees of freedom lesser than critical value

it can be calculated in excel as =+T.DIST.RT(5.8009,16)

p-value = 0.00014

e)

since p-value is lesser than 0.01 reject null hypothesis

Hence answer is Yes

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