Question

Determine the direction of the hypothesis test (one-sided left, one-sided right, bidirectional)
 Determine the test statistic (z* or t*) and the p-value for each of the following situations and
 Determine if they would cause the rejection of the null hypothesis if the confidence level was set
at 95% in each case. (Hint: be wary of the sample size) [2 points each]:
a) Ho: μ = 50 mL, Ha: μ ≠ 50 mL, sample mean = 48.1 mL, sample standard deviation = 5, n = 40
b) Ho: μ ≤ 8.4 m3
, Ha: μ > 8.4 m3
, sample mean = 10 m3
, s = 3.5, n = 25
c) Ho: μ ≥ 20oC, Ha: μ < 20oC , sample mean = 17.1oC, s =4.6
oC, n = 12
d) Ho: μ = 357 s, Ha: μ ≠ 380 s, sample mean = 410 s, s = 75, n = 40
e) Ho: μ ≤ 46 units, Ha: μ > 46 units, sample mean = 50 units, s = 9.5, n = 41

Answer 1

Answer on Question #60615 – Math – Statistics and Probability

Question

- Determine the direction of the hypothesis test (one-sided left, one-sided right, bidirectional)

- Determine the test statistic ($z^*$ or $t^*$) and the p-value for each of the following situations and

- Determine if they would cause the rejection of the null hypothesis if the confidence level was set at 95% in each case. (Hint: be wary of the sample size) [2 points each]:

a) Ho: $\mu = 50 \text{ mL}$, Ha: $\mu \neq 50 \text{ mL}$, sample mean = 48.1 mL, sample standard deviation = 5, n = 40;

b) Ho: $\mu \leq 8.4 \text{ mL}$, Ha: $\mu > 8.4 \text{ mL}$, sample mean = 10 mL, s = 3.5 mL, n = 25;

c) Ho: $\mu \geq 20{}^{\circ} \text{C}$, Ha: $\mu < 20{}^{\circ} \text{C}$, sample mean = $17.1{}^{\circ} \text{C}$, s = $4.6{}^{\circ} \text{C}$, n = 12;

d) Ho: $\mu = 357 \text{ s}$, Ha: $\mu \neq 380 \text{ s}$, sample mean = 410 s, s = 75, n = 40;

e) Ho: $\mu \leq 46$ units, Ha: $\mu > 46$ units, sample mean = 50 units, s = 9.5, n = 41.

Solution

The general rule for when to use a $t^*$ statistic is when our sample size meets the following two requirements:

- The sample size is below 30

- The population standard deviation is unknown (estimated from your sample data)

In all our cases we know only sample standard deviation and therefore should use $t^*$ statistic.

a) Bidirectional; Test statistic $t^*$:

$$t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} = \frac{48.1 - 50}{\frac{s}{\sqrt{40}}} = -2.40.$$

$p = 0.021 < 0.05$. Reject the null hypothesis.

b) One-sided right; Test statistic $t^*$:

$$t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} = \frac{10 - 8.4}{\frac{3.5}{\sqrt{25}}} = 2.29.$$

$p = 0.016 < 0.05$. Reject the null hypothesis.

c) One-sided left; Test statistic $t^*$:

$$t = \frac {\bar {x} - \mu}{\frac {\bar {s}}{\sqrt {n}}} = \frac {17.1 - 20}{\frac {4.6}{\sqrt {12}}} = -2.18.$$

$p = 0.026 < 0.05$. Reject the null hypothesis.

d) Bidirectional; Test statistic $t^*$:

$$t = \frac {\bar {x} - \mu}{\frac {\bar {s}}{\sqrt {n}}} = \frac {410 - 380}{\frac {75}{\sqrt {40}}} = 2.53.$$

$p = 0.016 < 0.05$. Reject the null hypothesis.

e) One-sided right; Test statistic $t^*$:

$$t = \frac {\bar {x} - \mu}{\frac {\bar {s}}{\sqrt {n}}} = \frac {50 - 46}{\frac {9.5}{\sqrt {41}}} = 2.70.$$

$p = 0.005 < 0.05$. Reject the null hypothesis.

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