Question

According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? Why or why not? Test at the 5% level.

In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Nationally 1 in 88 children are diagnosed with ASD ("CDC features -," 2013).
Is there sufficient data to show that the incident of ASD is more in Arizona than nationally? Why or why not? Test at the 1% level.

Answer 1

Solution:

1) The null and alternative hypotheses are as follows:

H_{​​​​​​0} : P = 23% = 0.23 i.e.The population proportion of identity theft in Alaska is equal to 0.23.

H_{​​​​​​1} : P < 0.23 i.e.The population proportion of identity theft in Alaska is less than 0.23.

To test the hypothesis we shall use z-test for single proportion. The test statistic is given as follows:

Where, p is sample proportion, P is population proportion specified under H_{​​​​​​0}, Q = 1 - P and n is sample size.

Sample proportion of the identity theft in Alaska is,

p = 321/1432 = 0.2242

Sample size (n) = 1432,

P = 0.23 and Q = 1 - 0.23 = 0.77

The value of the test statistic is -0.52496.

Since, our test is left-tailed test, therefore we shall obtain left-tailed p-value for the test statistic. The left-tailed p-value is given as follows,

p-value = P(Z < value of the test statistic)

p-value = P(Z < -0.52496)

Using "pnorm" function of R we get, P(Z < -0.52496) = 0.2998

Hence, p-value = 0.2998

P-value of the test statistic is 0.2998.

We make decision rule as follows:

If p-value is greater than the significance level, then we fail to reject the null hypothesis (H_{​​​​​​0}) at given significance level.

If p-value is less than the significance level, then we reject the null hypothesis (H_{​​​​​​0}) at given significance level.

p-value = 0.2998 and significance level = 5% = 0.05

(0.2998 > 0.05)

Since, p-value is greater than the significance level of 0.05, therefore we shall be fail to reject the null hypothesis (H_{​​​​​​0}) at 0.05 significance level.

Conclusion : At 5% level of significance, there is not enough evidence to conclude that Alaska had a lower proportion of identity theft than 23%.