Lab 6: Solutions to questions 02 and 03
One Sample Z and T tests
Question 02
From previous studies, you believe the population standard deviation is 2.0.
(a)
Step 1: Assumptions
Population: College-aged adults.
The comparison distribution will be a distribution of means.
The hypothesis test will be a z test because we have only one sample and we know the population mean and standard deviation.
Step 2: State the null and alternative hypotheses.
Hypotheses are always about populations, not samples!
Null hypothesis: Population mean number hours of sleep for college students is 8.
\[H_0: \mu = 8\]
Alternative hypothesis: Population mean number hours of sleep for college students is different from 8.
\[H_a: \mu \neq 8\]
Step 3: Determine the characteristics of the comparison distribution
For z tests, we must know the mean and the standard error of the population of scores
From Question 01, the sample mean is 8.2921.
And the standard deviation is:
\[\sigma_\bar{X} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2}{\sqrt{95}} = 0.2051957\]
Step 4: Determine the significance level
The exercise already did that for us.
\[\alpha = 0.05\]
Step 5: Calculate Test Statistic
\[z = \dfrac{\bar{x} - \mu}{\sigma_{\bar{X}}} = \dfrac{8.2921 - 8}{0.205195} = 1.42\]
Step 6.1: Conclude (Statiscal way)
Because this is a two-sided test (alternative hypothesis: not equal), p-value is going to be:
\[P(|Z| > 1.42) = 2 P( Z > 1.42)\]
Therefore, p-value is:
## [1] 0.1556So, p-value > 0.05. Therefore, we do not reject the null hypothesis.
Step 6.2: Conclude (English)
There is enough (or significant) evidence to conclude that the population mean number hours of sleep for college students is different from 8.
(b)
To construct the 95% confidence interval we need z so that
\[P(- z < Z < z) = .95\]
Hence, z = 1.96.
Therefore, using the equation:
\[\bar{x} \pm z \times \sigma_\bar{X} = 8.2921 \pm 1.96 \times 0.2051957\]
Ergo,
\[CI 95\% = [7.89, 8.694]\]
Question 03
From previous studies, you believe the population standard deviation is 2.0.
(a)
We do not know the population standard deviation. Therefore, we have to use a T test.
(b)
Step 1: Assumptions
Population: College-aged adults.
The comparison distribution will be a distribution of means.
The hypothesis test will be a t test because we have only one sample and we do not know the population standard deviation.
Step 2: State the null and alternative hypotheses.
Does not change from Question 02 a!
Hypotheses are always about populations, not samples!
Null hypothesis: Population mean number hours of sleep for college students is 8.
\[H_0: \mu = 8\]
Alternative hypothesis: Population mean number hours of sleep for college students is different from 8.
\[H_a: \mu \neq 8\]
Step 3: Determine the characteristics of the comparison distribution
From Question 01, the sample mean is 8.2921.
And the standard deviation now is:
\[s_\bar{X} = \dfrac{s}{\sqrt{n}} = \dfrac{1.66}{\sqrt{95}} = 0.1703124\].
Therefore, the
Step 4: Determine the significance level
Does not change from Question 02 a!
The exercise already did that for us.
\[\alpha = 0.05\]
Step 5: Calculate Test Statistic
\[t = \dfrac{\bar{x} - \mu}{s_{\bar{X}}} = \dfrac{8.2921 - 8}{0.1703124} = 1.715084\].
Step 6.1: Conclude (Statiscal way)
Because this is a two-sided test (alternative hypothesis: not equal), p-value is going to be:
\[P(|T| > 1.71) = 2 P( Z > 1.71)\]
With degrees of freedom = 95 - 1 = 94.
Therefore, p-value is:
## [1] 0.09056So, p-value > 0.05. Therefore, we do not reject the null hypothesis.
Step 6.2: Conclude (English)
There is enough (or significant) evidence to conclude that the population mean number hours of sleep for college students is different from 8.
Doing in SPSS: Output
Same results!
(c)
To construct the 95% confidence interval we need t so that
\[P(- t < T < t) = .95\]
Hence, t = 1.99.
Therefore, using the equation:
\[\bar{x} \pm t \times s_\bar{X} = 8.2921 \pm 1.99 \times 0.1703124\]
Ergo,
\[CI 95\% = [7.95, 8.63].\]
Doing in SPSS: Output
\[CI 95\% = [8 + (-0.0464), 8 + 0.6303] \approx [7.95, 8.63]\]