# 【四技應用統計學】假設檢定綜合練習題2

5.A high school counselor wishes to see if the average number of dropounts in her school is 【21】. She reviews the last 【17】 years and finds that the number of dropounts each year is as shown. At α=【0.01, is the hypothesis refutable?

【12】 【18】 【24】【16】 【21】 【20】 【18】 【19】

【19】 【22】 【25】【16】【18 】【19】 【19】 【20】

【23】

6.A recent study in a small city stated that the average age of robbery victims was 【63.5】 years. A sample of【 20】 recent victims had a mean of 【63.7】 years and a standard deviation of 【1.9】. At α= 【0.05】, is the average a ge higher than originally believed? Use the P-value method.

7.A dietitian read in a survey that at least 【55%】 of adults do not eat breakfast at least 【3】 days a week. To verify this, she selected a random sample of【 80】 adults and asked them how many days a week they skipped breakfast . A total of【 50%】 responded that they skipped breakfast at least 【3】 days a week.At α=【0.10】, test the claim.

8.It has been hypothesized that standard deviation of the germination time of radish seeds is 【8】 days . The standard deviation of a sample of【 60】 radish plants' germination times was【 6 】days. At α= 【0.01】, test the claim.

Rating

5.

t-test statistic = -2.148, p-value= 0.047 > α(=0.01) ==> Do NOT reject the null hypothesis that the average number of dropounts is 21.

6.

t-test statistic = (63.7-63.5) / 1.9/SQRT(20) = 0.47075,

p-value= P(t19 > 0.47075) = 0.321589 > α(=0.05) ==> Do NOT reject the null hypothesis that the average age is 63.5.

7.

Z statistic = (0.5-0.55) / SQRT(0.55*0.45/80) = -0.898933

p-value= P(Z > -0.898933) = 0.815656 > α(=0.10) ==> Do NOT reject the null hypothesis that the less than or equal to 55% of adults do not eat breakfast at least 3 days a week.

8.

chi-square statistic = 59*6^2 / 8^2 = 33.1875,

p-value= 2*P(χ2 > 33.1875) = 0.005304 < α(=0.01) ==> REJECT the null hypothesis that the standard deviation of the germination time of radish seeds is 8 days.