Question

The data shown below represent the age (in weeks) at which babies first crawl, based on a survey of 12 mothers. Complete parts (a) through (c) below. 355 52 30 47 37 30 47 44 58 35 Click the icon to view the table of critical correlation coefficient values for normal probability plots (a) Draw a normal probability plot to determine this reasonable to conclude the data come from a population that is normally distributed. Choose the correct answer belon os o os the critical valve reasonable to conclude that the Since the correlation between the expected 2-scores and the observed data. data come from a population that is normally distributed (Round to three decimal places as needed.) (b) Draw a boxplot to check for outiers. Choose the correct answer below.

Answer 1

Here we can easily make scatter plot of theoritical and sample quantiles in R:

here is the code along with output:

x <- c(52, 30, 44, 35, 47, 37, 56, 26, 30, 47, 35, 26) #A) scatter <- qqnorm(x) Normal Q-Q Plot 55 50 0 0 45 Sample Quantiles 40 35 30 25 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Theoretical Quantiles cor (scatter$x, scatter$y) ## [1] 0.9717363

#B) boxplot(x, horizontal = T) 25 30 35 TTTT 40 45 50 55 #c) lower <- mean(x)-qt(0.975, 11)*sd(x)/sqrt(12); lower ## [1] 32.26411 upper <- mean(x)+qt(0.975, 11)*sd(x)/sqrt(12); upper ## [1] 45.23589

A)

AS we can see the above output that option B is correct.

Then we can compute the correlation between theoritical qunatiles and sample

and we see above that correlation between them is  0.9717363 which implies that their is a very strong correlation between them.

It implies that the data is there from normal distribution.

Correlation can also be computed manully by formula:

ΣΕ 1,9% - ny – ni2) ΣΕ1 – ny?) VIΣ

B)

From the above output we can see that Option B from boxplot is correct.

From the boxplot we can easily see that there are no observations beyond the wiskers (i.e., no points beyond 1.5 * IQR) of boxplot which clearly indicates that there is no outliers in the data.

option C is correct.

C)

Here we need to find the 95% COnfidence interval of mean of the data given:

we know that if population variance is not known then we will use t distribution.

$\bar{x} \pm t_{\alpha/2, n-1}s/\sqrt{n}$

where

$\bar{x}$ = 38.75

$t_{\alpha/2, n-1}$ is the critical value of t distribution with n-1 df at alpha level of significance  

  $s = \sqrt{\sum_{i=1}^{n}x_i^2-n\bar{x}^2} = 10.20806$

n = 12

Then Confidence interval is given by:

(32.26411, 45.23589)

So this confidence interval implies that probability of sample mean lies between (32.26411, 45.23589) is 0.95

So option B is correct.