The greatest common factor (GCF) of 40 and 44 is 4.

Explanation:

44 = 4 x 11

40 = 4 x 10

44 + 40 = (4 x 11) plus (4 x10) = 4 times (11+ 10)

4 0

3 years ago

Test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.

VikaD [51]

Answer:

1. H0 : p = 0.9

H1 : p ≠ 0.9

2. The test is two tailed.

3. Reject the null hypothesis

Step-by-step explanation:

We are given that we have to test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.

So, Null Hypothesis, $H_0$ : p = 0.90

Alternate Hypothesis, $H_1$ : p $\neq$ 0.9

Here, the test is two tailed because we have given that to test the claim that the proportion of people who own cats is significantly different than 90% which means it can be less than 0.90 or more than 0.90.

Now, test statistics is given by;

$\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1) , where, n = sample size = 100

$\hat p$ = 0.94 (given)

So, Test statistics = $\frac{0.94-0.90}{\sqrt{\frac{0.94(1-0.94)}{100} } }$ = 1.68

Now, P-value = P(Z > 1.68) = 1 - P(Z <= 1.68)

= 1 - 0.95352 = 0.0465 ≈ 0.05 or 5%

Now, our decision rule is that;

If p-value < significance level ⇒ Reject null hypothesis

If p-value > significance level ⇒ Accept null hypothesis

Since, here p-value is less than significance level as 0.05 < 0.1, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that proportion of people who own cats is significantly different than 90%.

8 0

3 years ago