Question

Suppose you are given the following sample from average body temperature of humans:
97.7
98.8
98.0
98.0
98.3
98.5
97.3
98.7
97.4
98.9
98.6
99.5
97.5
97.3
97.6
98.2
99.6
Use a 0.05 significance level to test the belief that population mean is less than 98.60F. Use P-value method, Critical value method and Confidence interval method to test the hypothesis.

Answer 1

∑x =    1669.9  

∑x² =    164041.73  

n =    17  

Mean , x̅ = Ʃx/n = 1669.9/17 = 98.2294

Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(164041.73-(1669.9)²/17)/(17-1)] = 0.7261

Null and Alternative hypothesis:  

Ho : µ = 98.6

H1 : µ <    98.6

Test statistic:  

t = (x̅- µ)/(s/√n) = (98.2294 - 98.6)/(0.7261/√17) =    -2.1044

df = n-1 =    16

p-value :  

p-value = T.DIST(-2.1044, 16, 1) =    0.0258

Decision:  

p-value < α, Reject the null hypothesis  

There is enough evidence to conclude that population mean is less than 98.60.

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Critical value :  

Left tailed critical value, t-crit = T.INV(0.05, 16) =    -1.746

Reject Ho if t < -1.746  

Decision:  

Reject the null hypothesis  

Conclusion:  

There is enough evidence to conclude that population mean is less than 98.60.

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At α = 0.05 and df = n-1 = 16, critical value, t-crit = ABS(T.INV(0.05, 16)) = 1.746

One-sided Upper confidence interval = x̅ + t-crit*s/√n = 98.2294 + 1.746 * 0.7261/√17 = 98.537

As the one-sided upper confidence interval is less than 98.6 so we reject the null hypothesis and conclude that population mean is less than 98.60.