LINK1)
Formulating the null and alternative hypotheses,

Ho:   u   =   32
Ha:    u   =/   32 [ANSWER, A]

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b)

Yes, becuse the mileage is assumed to be normally distributed. Also, the sample mean looks different from 32.

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c)

As we can see, this is a    two   tailed test.

Thus, getting the critical t,
df = n - 1 =    22
tcrit =    +/-   2.073873068   [ANSWER, CRITICAL T]

Getting the test statistic, as

X = sample mean =    30.15
uo = hypothesized mean =    32
n = sample size =    23
s = standard deviation =    4.87

Thus, t = (X - uo) * sqrt(n) / s =    -1.821825117 [ANSWER, T STATISTIC]

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As -2.074<t<2.074, we FAIL TO REJECT THE NULL HYPOTHESIS.

There is no significant evidence that the mean mpg is different to 32.

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d)

Formulating the null and alternative hypotheses,

Ho:   u   >=   32
Ha:    u   <   32

As we can see, this is a    left   tailed test.

Thus, getting the critical t,
df = n - 1 =    22
tcrit =    -1.717144374

Getting the test statistic, as

X = sample mean =    30.15
uo = hypothesized mean =    32
n = sample size =    23
s = standard deviation =    4.87

Thus, t = (X - uo) * sqrt(n) / s =    -1.821825117
As t < -.1717, WE REJECT THE NULL HYPOTHESIS.

There is significant evidence that the mean mpg is less than 32.


LINK2)
1. Ho: u=>130

Ha: u<130

2. t = (130-126)/15/sqrt(30) = 1.461

Critical value against 29 degrees of freedom is 1.699

since 1.461< 1.699 we fail to reject null hypothesis and conclude that there is not enough evidence to conclude that it is significantly lower than 130

3. Ho: u=>130

Ha: u<130

t = (130-126)/15/sqrt(100) = 2.667

Critical value against 99 degrees of freedom is 1.660

since 2.667> 1.660 we to reject null hypothesis and conclude that there enough evidence that wieght is significantly lower than 130.

4. The results are different because in part 3, the standard error has decreased from 15/sqrt(30) to 15/sqrt(100) which happened because the sample size increases