Question

The accompanying table lists body temperatures from 68 different randomly selected subjects measured at two different times in a day. Assume that the paired sample data are simple random samples and the differences have a distribution that is approximately normal. Complete parts (a) and (b) below. Click the icon to view the data on body temperatures. a. Use a 0.05 significance level to test the claim that there is no difference between body temperatures measured at 8 AM and at 12 AM In this example, is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the temperature at 8 AM minus the temperature at 12 AM. What are the null and alternative hypotheses for the hypothesis test? Ho Hd = 0°F H: 0 °F (Type integers or decimals. Do not round.) Identify the test statistic. t = -8.21 (Round to two decimal places as needed.) Identify the P-value. P-value = 0.000 (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? Since the P-value is less than the significance level, reject Ho. There is sufficient evidence to warrant rejection of the claim of no difference between body temperatures measured at 8 AM and at 12 AM b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)? The confidence interval is F< < °F.

Answer 1

Sample mean of the difference using excel function AVERAGE(), x̅_d = -0.8279

Sample standard deviation of the difference using excel function STDEV.S(), s_d = 0.8313

Sample size, n = 68

α = 0.05

a) Null and Alternative hypothesis:

Ho : µ_d = 0

H1 : µ_d ≠ 0

Test statistic:

t = (x̅_d)/(s_d/√n) = (-0.8279)/(0.8313/√68) = -8.21

df = n-1 = 67

p-value :

Two tailed p-value = T.DIST.2T(ABS(-8.2133), 67) = 0.0000

Conclusion:  

Since the p-value is less than the significance level, Reject Ho. There is sufficient evidence to warrant rejection of the claim of no difference between body temperature measured at 8 AM and at 12 AM.

b) 95% Confidence interval :

At α = 0.05 and df = n-1 = 67, two tailed critical value, t-crit = T.INV.2T(0.05, 67) = 1.996

Lower Bound = x̅d - t-crit*sd/√n = -0.8279 - 1.996 * 0.8313/√68 = -1.0291

Upper Bound = x̅d + t-crit*sd/√n = -0.8279 + 1.996 * 0.8313/√68 = -0.6267

-1.0291 < µ_d < -0.6267