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You may need to use the appropriate technology to answer this question. Consider the following hypothesis test. Hou 55 H: 4 <

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Answer #1

As the population standard deviation is not we will use t distribution in all test.

a.Test statistic:

t = \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{54-55}{\frac{5.2}{\sqrt{36}}}=-1.154

d.f. = n-1= 36-1= 35

P-value for one tailed t test with test statistic -1.154 wif d.f. (35) by Excel formula=TDIST(1.154,35,1)=0.128

As P-value(0.128) is greater than alpha(0.01) we fail to reject the null hypothesis.

So answer is 1596216893278_image.png

b.

t = \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{53-55}{\frac{4.7}{\sqrt{36}}}=-2.553

d.f.=36-1=35

P-value for one tailed t test with test statistic -2.553 with d.f. (35) by Excel formula=TDIST(2.553,35,1)=0.008

As P-value(0.008) is less than alpha(0.01) we reject the null hypothesis.

So the answer is 1596217254208_image.png

c

t = \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{56-55}{\frac{6}{\sqrt{36}}}=1

P-value for one tailed t test with test statistic 1 with d.f. (35) by Excel formula=TDIST(1,35,1)=0.162

As P-value(0.162) is greater than alpha(0.01) we fail to reject the null hypothesis.

so the answer is

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