Statistics 1060: Tests and Confidence Intervals

TEST AND CONFIDENCE INTERVALS FOR $\mu$ , ( $\sigma$ unknown)

n = 45, s = 0.6 sec, $\overline{x} = 7.6$ sec, and $\mu_{0} = 7.5$ sec.

$H_{0}: \ \mu = 7.5$ sec.

$H_{A}: \ \mu > 7.5$ sec.

Where mu is true mean time required to inflate a certain type of raft.

$t = \frac{\overline{x} - \mu_{0}}{\frac{s}{\sqrt{n}}} = \frac{7.6 - 7.5}{\frac{0.6}{\sqrt{45}}} = 1.118$

since n = 45 and t has 45 - 1 = 44 df

P-value = P(T > 1.118)

0.10 < p < 0.15

The value using the CDF command on the computer is 0.1348.

MTB > CDF.1118;

SUBC > t 44

1.118 0.8652

so 1 - 0.8652 = 0.1348

The p-value is large, indicating little evidence against the null hypothesis.

$\alpha = 0.10$ . Since p-value $> \alpha$ , do not reject H₀. On the basis of this sample there is not sufficient evidence to reject claim that the mean inflation time is 7.4 sec. The null hypothesis is not rejected at $\alpha = 0.10$ , $\alpha = 0.05$ or $\alpha = 0.01$ since in all three cases the p-value is larger than the specified $\alpha$ . Conclude that this data does not support the idea that the inflation is significantly lower than the specified time of 7.5 sec.

t has 44 df. Using the table, use either t with 40 df or t with 50 df.

For 40 df. t^* = 2.021.

For 50 df. t^* = 2.009.

Using the INVCDF command in MINITAB, for 44 df t^* = 2.015

MTB > INVCDF.025;

SUBC > t 44.

.0250 -2.1054

The $95\%$ confidence interval for $\mu$ is $7.6 \pm 2.015\frac{0.6}{\sqrt{45}} = 7.6 \pm 0.18$ .

(7.42 sec, 7.78 sec)

With t^* = 2.021 and t^* = 2.009 we also get (7.42 sec, 7.78 sec).

n = 10, s = 8.0 ml, $\overline{x} = 170 ml$ or $\mu_{0} = 175 ml$

$H_{0}: \ \mu = 175 ml$

$H_{A}: \ \mu \neq 175 ml$

where $\mu$ is the true mean amount dispensed per cup.

$t = \frac{\overline{x} - \mu_{0}}{\frac{s}{\sqrt{n}}} = \frac{170 - 175}{\frac{8.0}{\sqrt{10}}} = -1.98$

t has 9 df

p-value = $2P(t_{9 \ df} > 1.98)$

2(0.025 < p < 0.05)

0.05 < p < 0.10

Using the computer the p-value is 2(0.0395) = (0.079).

At $\alpha = 0.05$ the null hypothesis is not rejected since the p-value is greater than $\alpha$ . Conclude that on the basis of the evidence in this sample, the machine is not working properly so it is necessary to fix the machine.

At $\alpha = 0.10$ the null hypothesis is rejected since the p-value is less than $\alpha$ . Conclude the machine is not working properly. The problem seems to be underfilling.

t has n-1 = 10 - 1 = 9 df.

For a $95\%$ confidence interval, t^* = 2.262.

$170 \pm 2.262\frac{8.0}{\sqrt{10}} = 170 \pm 5.72$

The $95\%$ confidence interval for $\mu$ is (164.3 ml, 175.8 ml)

For a $90\%$ confidence interval, t^* = 1.833.

$170 \pm 1.833\frac{8.0}{\sqrt{10}} = 170 \pm 4.64$

The $90\%$ confidence interval for $\mu$ is (165.4 ml, 174.6 ml)

The Relationship Between Confidence Intervals and Tests

The $95\%$ confidence interval for $\mu$ is (164.3 ml, 174.6 ml). If we wish to test $H_{0}: \ \mu = 175 ml$ vs $H_{A}: \ \mu \neq 175 ml$ at $\alpha = 0.05$ , we can use the information in the confidence interval to predict the result of the test since we are operating at the same level of $\alpha$ in both cases (i.e. $\alpha = 0.05$ and 1 - C = 0.05) and the test is a two sided test. The confidence interval gives a list of the plausible values for $\mu$ . The value 175 ml ( $\mu_{0}$ ) is contained in $95\%$ the confidence interval so the null hypothesis is not rejected at $\alpha = 0.05$ .

For the same reasons at $\alpha = 0.10$ the null hypothesis is rejected since the $90\%$ confidence interval does not contain the value 175 ml ( $\mu_{0}$ ).

In addition to the usual assumption that the sample is random, since n is small (less than 30) we make the additional assumption that the distribution of a cupfull is approximately normal.

n = 3, s = 36.3 hours, $\overline{x} = 1173.6$ hours or $\mu_{0} = 1100$ hours

$H_{0}: \ \mu = 1100$ hours

$H_{A}: \ \mu > 1100$ hours

where $\mu$ is the true mean lifetime of the component

$t = \frac{\overline{x} - \mu_{0}}{\frac{s}{\sqrt{n}}} = \frac{1173.6 - 1100}{\frac{36.3}{\sqrt{3}}} = 3.51$

t has 2 df

p-value = $P(t_{2 \ df} > 3.51)$

0.025 < p < 0.05

Using the computer the p-value is p = 0.0362)

In this case the consequences of a Type I ( $\alpha$ ) error are that the component when the mean lifetime is not long enough. This could lead to failure of the mission.

The consequences of a Type II ( $\beta$ ) error are that the component is not used when the mean lifetime is long enough.

Jonathan Payne
1999-03-22