I am trying to solve a question but stuck with the steps. I can not find any similar questions. With help of some online resources to calculate some parts of the question but I can see that is not enough. I know my approach has lack of information but, this is the only thing I have reached, I was covid ill at the class hours and can not follow the class examples, I thought someone can help me to solve and learn the subject.
With help of the answers from here I try to give an answer. Still need some improvements but tried to do my best. I still do not have answer for question D and confused about CL(the part C) and Significance level(part B)
My answers:
$N\ =\ 9\ \ \ \ \ \ \ \ Sum\ of\ x\ =\ 3970\ \ \ \ \ \ \ \ Mean,\ µ = 441.1111 \ Variance,σ^2 = 161.1111$
$ \sigma\ =\ \sqrt{161.1111} = 12.6929$
$t\ =\ \frac{m\ -\ \mu}{s\ /\ \sqrt n}$
$t\ =\ \frac{500\ -\ 441.1111}{12.6929\ /\ \sqrt9} = 13.918545$
We subtract 1 to get degrees free 9 – 1 = 8
Degrees of freedom = n – 1 = 8
$Probability: P( T ≤ 13.918545) = 0.00000069 $ So, this is the p-Value
$We\ will\ reject\ H_0\ at \ \alpha = 1% $ and also any > 1%
$$ (i) 0.10\
The\ information\ from\ the\ first\ question\, the\ critical\ t-value\ for\ α = 0.10\ and\ df = 8,\ t_c=\ 1.86
\\ CI\ =\ (\bar{X}\ -\ \frac{t_c\ \times\ s}{\sqrt n},\ \bar{\ X}\ +\ \frac{t_c\ \times\ s}{\sqrt n})
\\CI\ =\ (441.1111\ -\ \frac{1.86\ \times\ 12.6929}{\sqrt9},\ \ 441.1111+\frac{1.86\ \times\ 12.6929}{\sqrt9})
= (441.1111 – 7.868, 441.1111 + 7.868)
= (433.243, 448.979)
$$
$For\ the\ other\ t_c\ values:$
(ii) 0.05 $t_c=\ 2.306$ CL = (431.354, 450.868) (iii) 0.01 $t_c=\
> 3.355$
= (426.915, 455.308)
Based on the answers in part 2 for (i) = 0.90, (ii) = 0.95, (iii) = 0.99, none of the confidence intervals contain 500.
The Question:
The worker says that the mean purchasing cost is 500 USD. We decide to test this.
For a random sample of 9 purchases drawn from a normally distributed population with unknown variance, the costs are:
430, 450, 450, 440, 460, 420, 430, 450, 440.
A) Conduct a hypothesis test of whether the population mean purchasing equals 500 USD. Include all assumptions, the hypotheses, test statistic, and P-value and interpret the result in context.
B) For which significance levels can you reject $H_0?$ (i) 0.10, (ii) 0.05, or (iii)
0.01.
C) Based on the answers in part B), for which confidence levels would the confidence interval contain 500? (i) 0.90, (ii) 0.95, or (iii) 0.99.
D) Use part B) and part C) to illustrate the correspondence between results
of significance tests and results of confidence intervals.
Best Answer
I will give some formulas, which may be proved by standard methods.
We have $H_0: X_i \sim N(a, \sigma^2)$,
$a= 500$, $n=9$, $\overline{X} = \frac{\sum_{i=1}^n X_i}{n} = 441.1111...$, $s^2 = \frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n} = \hat{\sigma^2}\frac{n-1}{n}$, $ \hat{\sigma^2} =\frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n-1} = 161.1111...$, $\hat{\sigma} = 12.69296...$.
We know that $\xi_n = \frac{\sqrt{n}(\overline{X} - a)}{\sigma} \sim N(0,1)$, $\eta_n = \frac{ns^2}{\sigma^2} \sim \chi_{n-1}^2$, $\xi_n$ and $\eta_n$ are independent, $$t = \frac{\xi_n}{\sqrt{\frac{\eta_n}{n-1} }} = \sqrt{n-1}\frac{ \overline{X} - a}{s} = \sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} \sim T_{n-1}.$$ and also $-t \sim T_{n-1}$, because $T_{n-1}$ is symmetric. Here $T_{n-1}$ is Student's distribution.
We have $t =\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} = -13.9185...$.
If $H_0$ is true then $|t| < u_{\frac{1+\gamma}2}$ with probability $\gamma$, where $u_c$ is a quantile function at $c$.
Put $\gamma = 0.99$. Hence $\frac{1+\gamma}2 = 0.995$. We know that $u_{0.995} = 5.04305...$ and hence $|t| < 5.04305...$ with probability $0.99$, if $H_0$ is true.
Hence we reject $H_0$ for all significance levels in $B$.
The confidence interval for $a$ we get form a condition $$|\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}.$$
It has the form: $$ a \in (\overline{X} - \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma} , \overline{X} + \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma}).$$
Even in case $\gamma = 0.99$, when the confidence interval is wider, than in cases $\gamma = 0.95$ and $\gamma = 0.90$, we saw that the condition $$|\sqrt{n}\frac{ \overline{X} - 500}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}$$ doesn't hold and hence the confinedce interval doesn't contain $a=500$.
So in $C)$ we get that $500$ is not contained is confinedce interval in any case.
D) We see that $H_0$ is accepted (significance level is fixed) if and only if $a$ is contained in a confidence interval of correspoding confidence level. The correspondence is illustrated.
Addition about D.
$H_0: X_i \sim N(a, \sigma^2)$.
Test has significanse level $1-\gamma$ (in other words, test has level of confidence $\gamma$). Test says that we shoud accept $H_0$ if and only if $$|\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}.$$
The confidence interval has the form: $$ a \in (\overline{X} - \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma} , \overline{X} + \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma}).$$
So the condition "$a$ belongs to the $\gamma$-confidence interval" is equivalent to condition $$|\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}$$
which is a necessary and sufficient condition to accept $H_0: X_i \sim N(a, \sigma^2)$ with significanse level $1-\gamma$ .
Conclusion. "$a$ belongs to the confidence interval (with level of condidence $\gamma$)" if and only if we accept $H_0: X_i \sim N(a, \sigma^2)$, using the test with significanse level $1-\gamma$ .