Question

An electronic retailer believes that, at most 40% of their cellphone inventory was sold during November. The retailer surveyed 80 dealers and found that 38% of the inventory was sold. Since 38% less than 40%, is the difference of 2 percentage points sampling error or a significant difference? Test at .05 level.

Answer 1

To determine whether the difference of 2 percentage points between the observed sales of 38% and the expected sales of 40% is a sampling error or a significant difference, we can conduct a hypothesis test at a significance level of 0.05.

Step-by-step explanation:

The question asks whether the difference of 2 percentage points between the observed sales of 38% and the expected sales of 40% is a sampling error or a significant difference. To evaluate this, we can conduct a hypothesis test at a significance level of 0.05.

Null hypothesis (H0): The proportion of cellphone inventory sold during November is equal to 40%.

Alternative hypothesis (H1): The proportion of cellphone inventory sold during November is less than 40%.

To perform the test, we use the sample proportion of 38% and the sample size of 80 dealers. We calculate the test statistic using the formula:

test statistic = (sample proportion - hypothesized proportion) / sqrt((hypothesized proportion * (1 - hypothesized proportion)) / sample size)

Using the test statistic, we can calculate the p-value by comparing it to the critical value from the t-distribution with (sample size - 1) degrees of freedom.

If the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that there is a significant difference. Otherwise, we fail to reject the null hypothesis and conclude that the difference is due to sampling error.

Answer 2

The difference of 2 percentage points may be due to sampling error

Step-by-step explanation:

Given that an electronic retailer believes that, at most 40% of their cellphone inventory was sold during November.

Sample size n =80

Sample proportion p = 0.38

`H_0: p=0.40\\H_a: p <0.40`

(left tailed test at 5% level)

p difference = -0.02

Assuming H0 to be true std error

=
`\sqrt{(pq)/(n) } \\=0.0548`

Z statistic = p diff/std error = -0.365

p value = 0.3575

Since p >0.05 we accept H0.

The difference of 2 percentage points may be due to sampling error