Question

what would Ho and Ha be for part b and how do i do part c

Proper digits of your student number before you start in e) QUESTION 6 Replace the symbols with proper digits of your student number before A statistician took a walk down the street toon durine d h e schools were in session. She recorded the ce and measured the height of the first children of different ares she encounter on this street. The oldest chi was a six-year old The statistician made sure that all the children were unrelated to each other. Here are the measurements Child T 2 Age (years) 2 Height (cm) 87 112 110 SSX-10, Sy-516.8, SSxy-68. Ex-20, Xy-501, Exy-2072 a) Find the equation describing the linear relationship between age and height for 2-to 6-year old children on that street Slope - 55x4/55x = e = 1.8 intercept = y-b, X = 100.0 - (6.8.4) [{height) = 73 Y = 1.84E-73 b) Conduct a test to determine if age, X, is useful linear predictor of height change. Assumes 4.2 and use a -0.01. T-test Therefore we do not d.f=3- ieri T< critical value S = 4.2 = 0.01 reject Ho at a=0.01 t=5.049 critical value.l= 5.841 Ho: c) Find the 95% confidence limits on the expected height of A-year-old child on the street. A is the last digit then use A-3,5 of your student number divided by 2 or 4.5 is zero (For example if the last digit is A = 9+2 = 4.5

Answer 1

Let the regression model that we are estimating be

$y=\beta_0+\beta_1x+\text{error}$

where y=height (cm)

x=Age (years)

We know the following

$\begin{align*} n&=5\\ \bar{x}&=\frac{\sum x}{n}=\frac{20}{5}=4\\ \bar{y}&=\frac{\sum y}{n}=\frac{501}{5}=100.2\\ SS_x&=10\\ SS_y&=516.8\\ SS_{xy}&=68\\ \end{align*}$

a) An estimate of the slope is

B1 SSry 68 SS10 = 6.8 SS

An estimate of the intercept is

$\begin{align*} \hat{\beta}_0=\bar{y}-\hat{\beta}_1\bar{x}=100.2-6.8\times 4=73 \end{align*}$

ans: The equation describing the linear relationship between age (x) and height (y) for 2-6 year old children on the street is

y = 73 + 6.8.1

b) Age would be a useful linear predictor of height change if the slope coefficient $\beta_1$ is not equal to zero.

The hypotheses are

$\begin{align*} &H_0:\beta_0=0\\ &H_a:\beta_1\ne 0 \end{align*}$

We know following

is the estimate of slope

B1 = 6.8

The estimated standard error of slope is

$\begin{align*} s_{\hat{\beta}_1}=\frac{s}{\sqrt{SS_x}}=\frac{4.2}{\sqrt{10}}=1.3282 \end{align*}$

The hypothesized value of the slope coefficient is $\begin{align*} \beta_{1H_0}=0 \end{align*}$

The test statistic is

$\begin{align*} t=\frac{\hat{\beta}_1-\beta_{1H_0}}{s_{\hat{\beta}_1}}=\frac{6.8-0}{1.3282}=5.120 \end{align*}$  

Note: We have used s=4.2 in the above calculations. you would get t=5.049 only if you use an unrounded value of s.

This is a 2 tailed test (The alternative hypothesis has "not equal to")

The right tail critical value for $\alpha=0.01$ is

$P(T>t_{\alpha/2})=\alpha/2=0.01/2=0.005$

The degrees of freedom are n-2=5-2=3

Using the t tables, for df=3, and the area under the right tail=0.005, we get $t_{\alpha/2}=5.841$

The critical values are -5.841, 5.841 (Note: 2 tail tests always have 2 critical values)

We will reject the null hypothesis if the test statistic does not lie within the acceptance region -5.841 to  5.841

Here, the test statistic is 5.120 and it lies within -5.841 to  5.841. Hence we do not reject the null hypothesis.

ans: We can conclude that there is no sufficient evidence to support the claim that age,x is useful linear predictor of height change.

c) The expected value of y (height) for x=A=4.5 is

$\begin{align*} \hat{y}=73+6.8\times 4.5=103.6 \end{align*}$

The standard error of expected value of y for x=4.5 is

$\begin{align*} s_{\hat{y}}=s\sqrt{\frac{1}{n}+\frac{(x-\bar{x})^2}{SS_x}}=4.2\times \sqrt{\frac{1}{5}+\frac{(4.5-4)^2}{10}}=1.9922 \end{align*}$

The significance level for 95% confidence level is $\alpha=1-95/100=0.05$

The right tail critical value is

$P(T>t_{\alpha/2})=\alpha/2=0.05/2=0.025$

Uing the t tables for df=3 and the area under the right tail=0.025, we get $t_{\alpha/2}=3.182$

The 95% confidence interval is

$\begin{align*} &\hat{y}\pm t_{\alpha/2}s_{\hat{y}}\\ \implies &103.6\pm 3.182\times 1.9922\\ \implies &[97.2607, 109.9393] \end{align*}$

ans: The 95% confidene interval on the expected height of a 4.5 year old child on the street is (97.2607, 109.9393) cm