Answer:
No, we can't reject the dealer's claim with a significance level of 0.05.
Step-by-step explanation:
We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.
You take a random sample of 600 cars in San Francisco and find that 141 are blue.
<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>
SO, Null Hypothesis,
: p = 25% {means that 25% of all cars in San Francisco are blue}
Alternate Hypothesis,
: p
25% {means that % of all cars in San Francisco who are blue is different from 25%}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. =
~ N(0,1)
where,
= sample proportion of 600 cars in San Francisco who are blue =
= 0.235
n = sample of cars = 600
So, <u><em>test statistics</em></u> = ![\frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }](https://tex.z-dn.net/?f=%5Cfrac%7B0.235-0.25%7D%7B%7B%5Csqrt%7B%5Cfrac%7B0.235%281-0.235%29%7D%7B600%7D%20%7D%20%7D%20%7D%20%7D)
= -0.866
<em>Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>
Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.