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This has me puzzled. I am doing a right sided t-test where the null hypothesis is rejected based on the critical t value and the p value but the sample confidence interval includes the expected value. There is something missing in my understanding because I thought if the confidence interval included the value you could conclude there is no difference between the sample and hypothesized value.

The null hypothesis is rejected for the following reasons:

  • Reject the null hypothesis because the t-statistic, 1.54, was greater than the critical t value, 1.383.

  • Reject the null hypothesis because the p-value, 0.079, was less than the significance level, 0.1.

Here are the values used in the calculation:

Ho: mean <= 50.5
Ha: mean > 50.5

Expected Mean: 50.5
Sample Mean: 53.7
Sample Standard Deviation: 6.567
Standard Error: 2.077
Sample Size: 10
Significance Level: .1

Sample 90% Confidence Interval: [49.89, 57.51]
p-Value: .079
t-Statistic: 1.54
Cohen's D: .49

Critical T: 1.383

Do any of you know why this is happening?

Thanks,

Kevin

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  • $\begingroup$ Your basic idea that a 90% confidence interval contains a hypothetical value not rejected at the 10% significance level is right. But I think you are mixing a one-sided test with a two-sided confidence interval. $\endgroup$ – BruceET Jan 8 at 10:29
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I think you may be using a two-sided 90% confidence interval along with a one-sided test at the 10% level. If so, your result is not surprising.

Your data must have been something like my 10-vector y below. [Computations in R.]

y
[1] 51.5 48.3 49.7 54.8 60.3 53.7 51.3 48.8 49.3 69.3
mean(y);  sd(y)
[1] 53.7          # sample mean
[1] 6.559641      # sample standard deviation

A (one-sided) t.test of $H_0: \mu \le 50.5$ vs. $H_a: \mu > 50.5$ rejects at the 10% level because the P-value $0.079 < 0.10 = 10\%.$

t.test(y, mu=50.5, alt="greater", conf.lev=.9)

        One Sample t-test

data:  y
t = 1.5427, df = 9, p-value = 0.07865
alternative hypothesis: true mean is greater than 50.5
90 percent confidence interval:
 50.83113      Inf
sample estimates:
mean of x 
     53.7 

The one-sided 90% confidence interval amounts to a lower bound of 50.83, which does not include the hypothetical value 50.5. This is appropriate because $H_0$ is rejected.

However, a two-sided test of $H_0: \mu = 50.5$ against $H_a: \mu \ne 50.5$ is not rejected at the 10% level because the P-value $0.1573 > .1 = 10\%.$ Accordingly, the corresponding two-sided 90% confidence interval, $(49.8975, 57.5025),$ does contain the hypothetical value $\mu = 50.5.$

t.test(y, mu=50.5, conf.lev=.9)

        One Sample t-test

data:  y
t = 1.5427, df = 9, p-value = 0.1573
alternative hypothesis: true mean is not equal to 50.5
90 percent confidence interval:
  49.8975 57.5025
sample estimates:
mean of x 
     53.7 
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