The score x which is a random variable with mean u and variance 80. Sample 64 x

Question

The score x which is a random variable with mean u and variance 80. Sample 64 x bar : 589 . Mean Sat score of two years was 570. Has avg increased over two years use .05 level of test & Avg length of trout claim is 19 inches, random sample of 15 fish caught mean length was 18.5 with sample standard deviation of 3.2 inches . Does data show abg length is less than 19 inches at 5% level of significance ? The score x which is a random variable with mean u and variance 80. Sample 64 x bar : 589 . Mean Sat score of two years was 570. Has avg increased over two years use .05 level of test & Avg length of trout claim is 19 inches, random sample of 15 fish caught mean length was 18.5 with sample standard deviation of 3.2 inches . Does data show abg length is less than 19 inches at 5% level of significance ? & Avg length of trout claim is 19 inches, random sample of 15 fish caught mean length was 18.5 with sample standard deviation of 3.2 inches . Does data show abg length is less than 19 inches at 5% level of significance ?

Explanation / Answer

Avg length of trout claim is 19 inches, random sample of 15 fish caught mean length was 18.5 with sample standard deviation of 3.2 inches . Does data show abg length is less than 19 inches at 5% level of significance ?

we know that xbar= 18.5 , mu = 19 and sd = 3.2 , n = 15

t = xbar-mu/(sd/sqrt(n))

so 18.5-19/(3.2/sqrt(15)) = -0.6051

now we calculate the t critcal from the t table for df = n-1 = 14 and alpha = 0.05

please note that this is a 1 tail test as we are interested in the "less than" part

so t critcal is 1.761 , now as our t stat is less than t critical , hence we reject null hypothesis in favor of alternate hypothesis and conclude that avg length is less than 19 inches

Ho : avg length is equal to19 inches

H1 : avg length is less than 19 inches

The score x which is a random variable with mean u and variance 80. Sample 64 x bar : 589 . Mean Sat score of two years was 570. Has avg increased over two years use .05 level of test

again using the same formula

we know that xbar= 589 , mu = 570 and sd = sqrt(80) , n = 64

t = xbar-mu/(sd/sqrt(n))

(589-570)/(8.944/8) = 16.99

now the t critical for df = 64-1 = 63 and alpha = 0.05

the critical value for 1 tail is 1.669 , as this is a directional test (increased over the years -- part of the question )

here

as t calulated > t critical , hence we fail to reject the null hypothesis

Ho : avg did not increased over two years

H1 : avg increased over two years

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