Can someone help please?

In a random sample of cars driven at low altitudes, of them exceeded a standard of grams of particulate pollution per gallon of

Orlov [11]

Complete question is;

In a random sample of 370 cars driven at low altitudes, 43 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed. In an independent random sample of 80 cars driven at high altitudes, 23 of them exceeded the standard. Can you conclude that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard at an level of significance? Group of answer choices

Answer:

Yes we can conclude that there is enough evidence to support the claim that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard (P-value = 0.00005).

Step-by-step explanation:

This is a hypothesis test for the difference between the proportions.

The claim is that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.

Then, the null and alternative hypothesis are:

H0 ; π1 - π2 = 0

H1 ; π1 - π2 < 0

The significance level would be established in 0.01.

The random sample 1 (low altitudes), of size n1 = 370 has a proportion of;

p1 = x1/n1

p1 = 43/370

p1 = 0.116

The random sample 2 (high altitudes), of size n2 = 80 has a proportion of;

p2 = x2/n2

p2 = 23/80

p2 = 0.288

The difference between proportions is pd = (p1-p2);

pd = p1 - p2 = 0.116 - 0.288

pd = -0.171

The pooled proportion, we need to calculate the standard error, is:

p = (x1 + x2)/(n1 + n2)

p = (43 + 23)/(370 + 80)

p = 66/450

p = 0.147

The estimated standard error of the difference between means is computed using the formula:

S_(p1-p2) = √[((p(1 - p)/n1) + ((p(1 - p)/n2)]

1 - p = 1 - 0.147 = 0.853

Thus;

S_(p1-p2) = √[((0.147 × 0.853)/370) + ((0.147 × 0.853)/80)]

S_(p1-p2) = 0.044

Now, we can use the formula for z-statistics as;

z = (pd - (π1 - π2))/S_(p1-p2)

z = (-0.171 - 0)/0.044

z = -3.89

Using z-distribution table, we have the p-value = 0.00005

Since the P-value of (0.00005) is smaller than the significance level (0.01), then the effect is significant.

We conclude that The null hypothesis is rejected.

Thus, there is enough evidence to support the claim that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.

6 0

3 years ago

A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11. Use the TI-84 Plus calculator. (a) I

sukhopar [10]

Answer:

a) We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

b) There is a 15.97% probability that x will be between 54 and 55.

c) The 47th percentile of x is $X = 52.877$ .

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11.

This means that $\mu = 53$ .