A school board administrator wishes to see if this year\'s average provincial Gr

Question

A school board administrator wishes to see if this year's average provincial Grade 3 test score is different from what it was in 2017. The average Grade 3 test score in 2017 was 63.54%. The administrator examines a random sample of 26 test scores from this year. The mean of this sample is 60.3% with a standard deviation of 1.22%.

Use this information and data to answer the following questions.

a. Select the appropriate null and alternative hypotheses used to test if if this year's average provincial Grade 3 test score is different from what it was in 2017.
A. H0:?=63.54%Ha:??63.54%H0:?=63.54%Ha:??63.54%
B. H0:??60.3%Ha:?=60.3%H0:??60.3%Ha:?=60.3%
C. H0:X?????63.54%Ha:X????=63.54%H0:X¯?63.54%Ha:X¯=63.54%
D. H0:??63.54%Ha:?=63.54%H0:??63.54%Ha:?=63.54%
E. H0:X????=60.3%Ha:X?????60.3%H0:X¯=60.3%Ha:X¯?60.3%
F. H0:X????=63.54%Ha:X?????63.54%H0:X¯=63.54%Ha:X¯?63.54%
G. H0:?=60.3%Ha:??60.3%H0:?=60.3%Ha:??60.3%
H. H0:X?????60.3%Ha:X????=60.3%H0:X¯?60.3%Ha:X¯=60.3%

b) Compute a 99% confidence interval for ??, this year's average provincial Grade 3 test score.
<?<

(round to 3 decimals)

c)Consider the hypotheses in part a. Based on the confidence interval, what decision do we come to with respect to our hypotheses?

A. Reject H0H0
B. Fail to reject H0H0,

Explanation / Answer

A. H0:?=63.54%, Ha:??63.54%

The formula for estimation is:

? = M ± t(sM)

where:

M = sample mean
t = t statistic determined by confidence level
sM = standard error = ?(s2/n)

M = 60.3
t = 2.79
sM = ?(1.222/26) = 0.24

? = M ± t(sM)
? = 60.3 ± 2.79*0.24
? = 60.3 ± 0.667

You can be 99% confident that the population mean (?) falls between 59.633 and 60.967.

As, it doesnt contain 63.54, null is rejected

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