1 - Null hypothesis: H0:
d = 0
Alternative hypothesis: Ha:
d < 0
2 - The test statistic is t = (
-
d) / (sd /
) = (-2.3 - 0) / (1.5 / √17) ≈ -4.47
3 - The critical value(s) is (are) t0 = t(0.05, 16) ≈ -1.746
4 - Since the test statistic is in the rejection region, we reject the null hypothesis. There is statistically significant evidence to reject the claim.
1: Null hypothesis: H0:
d = 0
Alternative hypothesis: Ha:
d < 0
2: Calculate the test statistic:
t = (
-
d) / (sd /
)
where
is the sample mean of the differences,
d is the population mean of the differences, sd is the standard deviation of the differences, and n is the sample size.
In this case,
= -2.3
d = 0
sd = 1.5
n = 17
t = (-2.3 - 0) / (1.5 / √17)
t ≈ -4.47
3:Find the critical value(s):
The critical value can be found using a t-distribution table or a statistical software. For a one-tailed test with α = 0.05 and degrees of freedom (df) = n - 1, the critical value is t0 = t(0.05, 16) ≈ -1.746
4: Make a decision:
Since the test statistic (-4.47) is in the rejection region (i.e., it is less than the critical value of -1.746), we reject the null hypothesis. This means that there is statistically significant evidence to reject the claim that the population mean of the differences is equal to zero in favor of the alternative hypothesis that it is less than zero.
The probable question may be:
Test the claim below about the mean of the differences for a population of paired data at the level of significance α. Assume the samples are random and dependent, and the populations are normally distributed.
Claim:
d≥0; α=0.05. Sample statistics: d=−2.3, sd=1.5, n=17
1 - Identify the null and alternative hypotheses.
2 - The test statistic is ? t = __
3 - The critical value(s) is (are) ? t0 = __
4 - Since the test statistic is __ (in/ not in) the rejection region, __ (reject/ fail to reject) the null hypothesis. There __ (is/ is not) statistically significant evidence to reject the claim.