Question

A life insurance representative believes that the mean age of people who buy their first life insurance plan is less than 35. To test his belief he takes a random sample of 15 customers who have just purchased their first life insurance. Their ages are 42, 43,28, 34, 30, 36, 25, 29, 32, 33, 27, 30, 22, 37, and 40. There is not enough evidence to say the data are nonnormal. Can we conclude at the 1% significance level that the insurance representative is correct?

Answer 1

Solution:

x	x2
42	1764
43	1849
28	784
34	1156
30	900
36	1296
25	625
29	841
32	1024
33	1089
27	729
30	900
22	484
37	1369
40	1600
---	---
$\sum$x=488	$\sum$x2=16410

The sample mean is $\bar x$

Mean $\bar x$   = ($\sum$x / n) )

=42+43+28+34+30+36+25+29+32+33+27+30+22+37+40 /15

=488 /15

=32.5333

Mean $\bar x$   =32.53

The sample standard is S

  S =$\sqrt{}$( $\sum$ x² ) - (( $\sum$ x)² / n ) n -1

=$\sqrt{}$(16410-(488)²15 )14

=$\sqrt{}$(16410-15876.2667)14

=$\sqrt{}$533.7333 /14

=$\sqrt{}$38.1238

=6.174

This is the left tailed test .

The null and alternative hypothesis is ,

H₀ :  $\mu$  = 35

H_a : $\mu$    $\neq$ 35

Test statistic = t

= ($\bar x$ - $\mu$ ) / s / $\sqrt$ n

= (35 - 32.53) / 6.17 / $\sqrt$ 15

= −1.550

Test statistic = t =−1.550

P-value =0.0717

$\alpha$ = 0.01

P-value ≥ $\alpha$

0.0717 ≥ 0.01

Do not reject the null hypothesis .

There is insufficient evidence to suggest that