Question

Consider the following hypotheses: H emptyset : mu = 70 H_{A} mu ne70 table or t table) Find the p-value for this test based on the following sample information. (You may find it useful to reference the appropriate table: z a. overline x =66: s =12.0: n = 25 O 0.02 ≤ p-value < 0.05 D - value > 0.1 O 0.01 s p-value < 0.02 O p-value < 0.01 O 0.05 s p-value < 0.1 b. overline x = 74 s = 120 n = 25 O p-value ≥ 0.10 0.05 s p-value < 0.1 O 0.02 s p-value < 0.05 O 0.01 s p-value < 0.02 O p-value <0.01 c. overline x = 67 ; s = 11.3 n = 25 O p-value > 0.10 O p-value < 0.01 O 0.02 ≤ p-value < 0.05 O 0.01 s p-value < 0.02 O 0.05 s p-value < 0.10

Answer 1

a.
`\(0.02 \leq \text{p-value} < 0.05\)`

b.
`\(\text{p-value} \geq 0.10\)`

c.
`\(\text{p-value} < 0.01\)`

Step-by-step explanation:

To find the p-values for the given hypotheses, a Z-test can be used to compare the sample mean
`(\(\overline{x}\))` to a known population mean
`(\(\mu\))` under the null hypothesis
`(\(H_0: \mu = 70\))`.

The formula to calculate the Z-test statistic is:

`\[ Z = \frac{\overline{x} - \mu}{(s)/(√(n))} \]`

a. For
`\(\overline{x} = 66\), \(s = 12.0\)`, and
`\(n = 25\)`:

\
`[ Z = (66 - 70)/((12.0)/(√(25))) = (-4)/(2.4) = -1.67 \]`

Using a standard normal distribution table or calculator, the corresponding two-tailed p-value is between 0.02 and 0.05.

b. For
`\(\overline{x} = 74\), \(s = 120\)`, and
`\(n = 25\)`:

`\[ Z = (74 - 70)/((120)/(√(25))) = (4)/(24) = 0.17 \]`

The p-value is greater than 0.10, indicating insufficient evidence to reject the null hypothesis.

c. For
`\(\overline{x} = 67\)`,
`\(s = 11.3\)`, and
`\(n = 25\)`:

`\[ Z = (67 - 70)/((11.3)/(√(25))) = (-3)/((11.3)/(5)) = -1.33 \]`

Using a standard normal distribution table or calculator, the corresponding two-tailed p-value is less than 0.01.

Therefore, the p-values for the given sample information fall into the ranges specified in options a, b, and c.