Assume the hypothesis $H_0 : u = 59$ by computing the relevant $p$-value

Question

An observed sample of four observations from $N(u, \sigma^2)$ distribution has mean $54.75$ and standard deviation $10.21$. A table of quantiles of the chisquare distribution is given below

$(a)$ Assume the hypothesis $H_0 : u = 59$ by computing the relevant $p$-value

(b) Compute a $0.9$ confidence interval for $u$

(c) Compute an equal tail $0.9$-confidence interval for $\sigma^2$

ans:

(a) $pvalue = P(Z \leq \frac{\bar{x} - H_0}{\sigma_{0}/\sqrt{n}})$

$pvalue = P(Z \leq \frac{54.75-59}{10.21/2})$ (why is $\sqrt{n} = 2$?)

$P(Z \leq -0.832517401)$ (not sure what to do next)

(b) $CI = (\bar{x} \pm t_{\frac{1+r}{2}} \frac{\sigma_{0}}{\sqrt{n}})$

$=(54.75 \pm t_{\frac{1+0.9}{2}} \frac{10.21}{2})$ not sure how to get the CI from this

(c)

$$\left(\frac{(n-1)\sigma_0^2}{x^2_{n-1}\frac{1+r}{2}}, \frac{(n-1)\sigma_0^2}{x^2_{n-1}\frac{1-r}{2}} \right)$$

$$=\left(\frac{3\cdot 10.21^2}{x^2_{3}\frac{1+0.95}{2}}, \frac{3\cdot 10.21^2}{x^2_{3}\frac{1-0.9}{2}} \right)$$

$$=\left(\frac{3.127323}{x^2_{3}(0.95)}, \frac{3.127323}{x^2_{3}(0.05)} \right)$$

not sure how to solve for $x^2_{3}(0.05)$ and $x^2_{3}(0.95)$

Any help appreciated!

Answer 1

In all parts you must use special tables (software functions, online calculators)!

a) $$p\text{-value} = P\left(t \leq \frac{54.75-59}{10.21/2}\right)=P(t_{df=3}\le -0.8325)=0.2331\\ \sqrt{n}=\sqrt{4}=2, \ \text{(because the sample size is $4$)}$$ Note: Use Student's t-table or Wolframalpha for: $P(t_{df=3}\le -0.8325)=0.2331$.

b) CC=$0.9$ (confidence coefficient). $$CI = \left(\bar{x} \pm t_{\frac{1\color{red}-CC}{2},n-1} \frac{s}{\sqrt{n}}\right)=\left(54.75 \pm t_{\frac{1\color{red}-0.9}{2},3} \frac{10.21}{2}\right)=\left(54.75 \pm t_{0.05,3} \frac{10.21}{2}\right)=\\ \left(54.75 - 2.3534\frac{10.21}{2},54.75 + 2.3534 \frac{10.21}{2}\right)= (42.74,66.76).$$ Note: Use Student's t-table or Wolframalpha for: $t_{0.05,3}=2.$

c) CC=$0.9$ (confidence coefficient). $$\left(\frac{(n-1)s^2}{\chi^2_{\frac{1-CC}{2},n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\frac{1-CC}{2},n-1}} \right)= \left(\frac{3\cdot 10.21^2}{\chi^2_{0.05,3}}, \frac{3\cdot 10.21^2}{\chi^2_{0.95,3}} \right)=\\ \left(\frac{3\cdot 10.21^2}{7.81}, \frac{3\cdot 10.21^2}{0.35} \right)=\left(40.04, 893.52 \right).$$ Note: Use Chi-squared table (or your post) or Wolframalpha for: $\chi^2_{0.05,3, right}=\chi^2_{0.95,3, left}=0.35$.

Answer 2

Comment: @Farruhola has provided numerical answers and some comments on mathods (+1). Because there seems to be some confusion about the CI for $\sigma^2,$ here is additional explanation for part (c). The data are normal, so you have $$Q = \frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(n-1 = 3).$$ Then find quantiles .05 and .95 of this chi-squared distribution, $L$ and $R,$ respectively, so that $L = 0.3519,\, U = 7.8147,$ according to the computation in R below. (Can you find these two numbers, to three places, in the table you provided?)

qchisq(c(.05, .95), 3)
## 0.3518463 7.8147279

Thus $P(L < Q < U) = 0.90.$ The 90% CI for $\sigma^2$ is found by solving the inequality in $L < Q < U$ to isolate $\sigma^2$ between bounds. Note that, after solving the inequality, $U$ is used for the lower bound and $L$ for the upper bound.

Check to make sure that the resulting numerical CI contains the point estimate $S^2$ of $\sigma^2.$ However, unlike the t confidence interval for $\mu,$ the point estimate is not at the exact center of the interval. (That is, the CI for $\sigma^2$ is not of the form "Point Estimate $\pm$ Margin of Error.")