Final answer:
The p-value for the two-tailed test given the sample mean of 26.76 is 0.0254. At the 0.05 significance level, we reject the null hypothesis, but at the 0.01 significance level, we do not reject it due to the higher p-value.
Step-by-step explanation:
Calculation of P-value and Decision on Null Hypothesis
To calculate the p-value, we use the given information: a population mean (μ) of 29, a sample mean (μ) of 26.76, a population standard deviation (σ) of 6, and a sample size (n) of 36. The test is two-tailed since the alternative hypothesis is μ ≠ 29.
The test statistic (z) is calculated using the formula: z = (μ - μ) / (σ / √n). Substituting the values, z = (26.76 - 29) / (6 / 6) = -2.24 / 1 = -2.24.
The p-value can be found by looking at the normal distribution table or using statistical software. For a z-score of -2.24, the p-value in each tail is approximately 0.0127. Since this is a two-tailed test, the total p-value is 2 × 0.0127 = 0.0254.
(a) p-value = 0.0254
(b) At a significance level (α) of 0.05, we reject the null hypothesis because the p-value is less than α (0.0254 < 0.05). This means the sample provides enough evidence to suggest a significant difference from the population mean of 29.
(c) At a significance level (α) of 0.01, we do not reject the null hypothesis because the p-value is greater than α (0.0254 > 0.01). We don't have sufficient evidence to conclude a significant difference at this level of significance.