**Question 1: **A study for 18 participants was considered to examine the ideal body weight in the population.

The
data is 107 119 99 114 120 104 88 114 124 116 101 121 152 100 128 114
95 117. (100 represents ideal body weight for the given problem)

Test whether the given data supports non-ideal body weight in the population. Given

$\frac{\sigma}{\sqrt{n}}$ = 3.40.

** Solution: **

Set the null and alternate hypotheses

Hypotheses: Null hypothesis $H_{0}$: $\mu$ =100

Alternative hypothesis $H_{1}$ :$\mu \neq$ 100 (Two sided)

Test Statistic:

As the data is assumed to be normal the Z test is given by:

$t_{cal}$ = $\frac{\bar x -\mu }{\frac{\sigma }{\sqrt{n}}}$

$\bar x$ = $\frac{(107 + 119 + 99 + 114 + 120 + 104 + 88 + 114 + 124 + 116 + 101 + 121 + 152 + 100 + 128 + 114 +95 + 117)}{18}$ = 112.94

$\mu$ = 100 (given)

$\frac{\sigma}{\sqrt{n}}$ = 3.40 (given)

$t_{cal}$ = $\frac{(112.94 - 100)}{3.40}$ = 3.81

p-value:

Under t tables for degrees of freedom (18 - 1 = 17) the $t_{tabu}$ is 3.81 and the p value is 0.0016( two-sided).

__Conclusion__:

From the above we see that there is good evidence to reject $H_{0}$ and we conclude the difference is significant.

**Question 2: **A research firm is interested to study the reaching time to campus for
the Cambridge university, as on average some people claim they take 30
minutes to reach from the parking lot. As many people are against the 30
minutes. Test the claim assuming $\bar x = 20$ and $\sigma = 6$. Level of significance is 0.10 and n = 5.

** Solution: **

The null and alternative hypotheses are

$H_{0} :\mu ≥ 30$

$H_{1} : \mu < 30$

Test statistic

$Z_{calc}$ = $\frac{\bar x - \mu _{0}}{\frac{\sigma }{\sqrt{n}}}$ = $\frac{20-30}{\frac{6}{\sqrt{5}}}$ = -3.727

Decision rule

Under Normal tables for $Z_{calc}$ = -3.727 we get $Z_{tabu}$ = -1.28 which is on the left in the tables.

__Conclusion__:

As
$z_{calc} < Z_{ tabu}$ our test statistic value lies in the
rejection region. So we reject the null hypothesis and conclude that
mean will be significantly less than 30, and now there is sufficient
evidence to prove the parking space will be less than 30.