### Chapter 7. Hypothesis Testing: One-sample t-test

### Confidence Interval for μ when σ is Unknown

Thus far, we have used the following formula to compute a #C\%\,CI# for a population mean #\mu#:

\[CI_{\mu}=\bigg(\bar{X} - z^*\cdot \cfrac{\sigma}{\sqrt{n}},\,\,\,\, \bar{X} + z^*\cdot \cfrac{\sigma}{\sqrt{n}} \bigg)\]

Just like the calculation of the #Z#-statistic, this calculation requires the population standard deviation #\sigma# to be known.

When #\sigma# is unknown, we will have to rely on the #t#*-distribution* and the *sample standard deviation #s# *to compute the confidence interval.

#\phantom{0}#

Confidence Interval for a Population Mean when σ is Unknown

Assuming the *sampling distribution of the sample mean *is (approximately) normal, the general formula for computing a #C\%# CI for a population mean #\mu#, based on a random sample of size #n#, when *σ* is unknown, is:

\[CI_{\mu}=\bigg(\bar{X} - t^*\cdot \cfrac{s}{\sqrt{n}},\,\,\,\, \bar{X} + t^*\cdot \cfrac{s}{\sqrt{n}} \bigg)\]

Where #t^*# is the *critical value *of the #t_{n-1}# distribution such that #\mathbb{P}(-t^* \leq t \leq t^*)=\frac{C}{100}#.

Calculating t* with Statistical Software

Let #C# be the *confidence level *in #\%#.

To calculate the *critical value* #t^*# in Excel, make use of the function **T.INV()**:

\[=\text{T.INV}((100+C)/200, n \text{ - } 1)\]

To calculate the *critical value* #t^*# in R, make use of the function **qt()**:

\[\text{qt}(p=(100+C)/200, df=n \text{ - } 1,lower.tail = \text{TRUE})\]

He collects a random sample of #78# students. On average, these students spent #7.5# hours a week on their homework with a standard deviation of #s=1.6# hours.

Construct a #90\%# confidence interval for the population mean #\mu#. Round your answers to #3# decimal places.

There are a number of different ways we can compute the *confidence interval*. Click on one of the panels to toggle a specific solution.

Since the *population standard deviation *#\sigma# is unknown, we will have to use the #t#-distribution and the *sample standard deviation* #s# to construct the confidence interval.

A sample size of #n=78# is considered large enough for the *Central Limit Theorem *to apply.

This means that, although the sample in question comes from a population having an unknown distribution, the *sampling distribution* *of the sample* *mean* is approximately normal.

Assuming the *sampling distribution of the sample mean *is (approximately) normal, the general formula for computing a #C\%\, CI# for the population mean #\mu#, based on a random sample of size #n#, is:

\[CI_{\mu}=\bigg(\bar{X} - t^*\cdot \cfrac{s}{\sqrt{n}},\,\,\,\, \bar{X} + t^*\cdot \cfrac{s}{\sqrt{n}} \bigg)\]

For a given *confidence level *#C# (in #\%#), the *critical value* #t^*# of the #t_{n-1}# is the value such that #\mathbb{P}(-t^* \leq t \leq t^*)=\cfrac{C}{100}#.

To calculate this critical value #t^*# in Excel, make use of the following function:

T.INV(probability, deg_freedom)

probability: A probability corresponding to the normal distribution.deg_freedom: The mean of the distribution.

Here, we have #C=90#. Thus, to calculate #t^*# such that #\mathbb{P}(-t^* \leq t \leq t^*)=0.90#, run the following command:

\[\begin{array}{c}

=\text{T.INV}((100+C)/200, n - 1)\\

\downarrow\\

=\text{T.INV}(190/200, 78 \text{ - } 1)

\end{array}\]

This gives:

\[t^* = 1.66488\]

Calculate the lower bound #L# of the confidence interval:

\[L = \bar{X} - t^* \cdot \cfrac{s}{\sqrt{n}} = 7.5 - 1.66488 \cdot \cfrac{1.6}{\sqrt{78}}=7.198\]

Calculate the lower bound #U# of the confidence interval:

\[U = \bar{X} + t^* \cdot \cfrac{s}{\sqrt{n}} = 7.5 + 1.66488 \cdot \cfrac{1.6}{\sqrt{78}}=7.802\]

Thus, the #90\%# confidence interval for the population mean #\mu# is:

\[CI_{\mu,\,90\%}=(7.198,\,\,\, 7.802)\]

Since the *population standard deviation *#\sigma# is unknown, we will have to use the #t#-distribution and the *sample standard deviation* #s# to construct the confidence interval.

A sample size of #n=78# is considered large enough for the *Central Limit Theorem *to apply.

This means that, although the sample in question comes from a population having an unknown distribution, the *sampling distribution* *of the sample* *mean* is approximately normal.

Assuming the *sampling distribution of the sample mean *is (approximately) normal, the general formula for computing a #C\%\, CI# for the population mean #\mu#, based on a random sample of size #n#, is:

\[CI_{\mu}=\bigg(\bar{X} - t^*\cdot \cfrac{s}{\sqrt{n}},\,\,\,\, \bar{X} + t^*\cdot \cfrac{s}{\sqrt{n}} \bigg)\]

For a given *confidence level *#C# (in #\%#), the *critical value* #t^*# of the #t_{n-1}# is the value such that #\mathbb{P}(-t^* \leq t \leq t^*)=\cfrac{C}{100}#.

To calculate this critical value #t^*# in R, make use of the following function:

qt(p, df, lower.tail)

p: A probability corresponding to the normal distribution.df: An integer indicating the number of degrees of freedom.lower.tail: If TRUE (default), probabilities are #\mathbb{P}(X \leq x)#, otherwise, #\mathbb{P}(X \gt x)#.

Here, we have #C=90#. Thus, to calculate #t^*# such that #\mathbb{P}(-t^* \leq t \leq t^*)=0.90#, run the following command:

\[\begin{array}{c}

\text{qt}(p = (100+C)/200, df = n \text{ - } 1, lower.tail = \text{TRUE})\\

\downarrow\\

\text{qt}(p =190/200, df = 78 \text { - } 1, lower.tail = \text{TRUE})

\end{array}\]

This gives:

\[t^* = 1.66488\]

Calculate the lower bound #L# of the confidence interval:

\[L = \bar{X} - t^* \cdot \cfrac{s}{\sqrt{n}} = 7.5 - 1.66488 \cdot \cfrac{1.6}{\sqrt{78}}=7.198\]

Calculate the lower bound #U# of the confidence interval:

\[U = \bar{X} + t^* \cdot \cfrac{s}{\sqrt{n}} = 7.5 + 1.66488 \cdot \cfrac{1.6}{\sqrt{78}}=7.802\]

Thus, the #90\%# confidence interval for the population mean #\mu# is:

\[CI_{\mu,\,90\%}=(7.198,\,\,\, 7.802)\]

#\phantom{0}#

Connection to Hypothesis Testing

There exists a direct connection between a *two-sided one-sample* #t#*-test* for #\mu# and a #(1-\alpha)\cdot 100\%# confidence interval for #\mu# based on the #t#-distribution:

- If #\mu_0# falls
*inside*the #(1 - \alpha)\cdot 100\%\,CI#, then #H_0: \mu=\mu_0# should not be rejected at the #\alpha# level of significance. - If #\mu_0# falls
*outside*of the #(1 - \alpha)\cdot 100\%\,CI#, then #H_0: \mu=\mu_0# should be rejected at the #\alpha# level of significance.

*#91\%#*confidence interval for a population mean #\mu#, computed based on a simple random sample from the population, is #(-8.039,\,\, -3.476)#.

Suppose you use the same sample to test #H_0: \mu = 0# against #H_a: \mu \neq 0# at the #\alpha = 0.09# level of significance.

What would be the conclusion?

Since the #91\%# confidence interval #(-8.039,\,\,-3.476)# does not contain the value #\mu_0 = 0#, we would reject #H_0: \mu = 0# at the #\alpha = 0.09# level of significance.

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