Question

1. A magazine reported 66% of all dog owners usually greet their dog before greeting their

Answer 1

Answer:

(a) There is 95% confidence that the true proportion of dog owners who greet their dogs first is between 0.475 and 0.775.

(b) The point estimate is 0.625. The margin of error is 0.15.

(c) It is plausible that the true proportion of all owners who greet their dog first is 66%.

Step-by-step explanation:

The (1 - α)% confidence interval for population parameter implies that there is a (1 - α) probability that the true value of the parameter is included in the interval. Or, the (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

In statistic, point estimation comprises of the use of sample data to estimate a distinct data value (known as a point estimate) which is to function as a "best guess" or "best estimate" of an unidentified population parameter. The point estimate of the population mean (µ) is the sample mean ($\bar x$).

The  (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error for this interval is:

$MOE=\frac{UL-LL}{2}$

For the hypothesis test, if the confidence interval consist of the null value of the parameter then the null hypothesis is accepted or else rejected.

The 95% confidence interval for the population proportion of owners who greet their dog first is,

CI = (0.475, 0.775)

(a)

The 95% confidence interval for the population proportion, (0.475, 0.775), implies that there is a 0.95 probability that the true proportion of dog owners who greet their dogs first is between 0.475 and 0.775.

Or, there is 95% confidence that the true proportion of dog owners who greet their dogs first is between 0.475 and 0.775.

(b)

The point estimate of the population proportion (p) is the sample mean ($\hat p$).

Compute the point estimate from the 95% confidence interval as follows:

Point estimate = (UL + LL)/2

                        $=\frac{0.775+0.475}{2}\\=0.625$

The point estimate is 0.625.

Compute the margin of error as follows:

$MOE=\frac{UL-LL}{2}=\frac{0.775-0.475}{2}=0.15$

The margin of error is 0.15.

(c)

The hypothesis to test whether the proportion of all owners who greet their dog first is 66% is:

H₀: The proportion of all owners who greet their dog first is 66%, i.e. p = 0.66

Hₐ: The proportion of all owners who greet their dog first is not 66%, i.e. p ≠ 0.66.

The 95% confidence interval for the population proportion of owners who greet their dog first is,

CI = (0.475, 0.775)

The 95% confidence interval consists of the null value, i.e. p = 0.66.

The null hypothesis was failed to rejected.

Thus, it is plausible that the true proportion of all owners who greet their dog first is 66%.