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Feb 20, 2024

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le‘i The manufacturer of a new chewing gum claims that at least 80% of dentists surveyed prefer their type of gum and recommend it for their patients who chew gum. An independent consumer research firm decides to test their claim. The findings in a sample of 200 dentists indicate that 74.1% of the respondents do actually prefer their gum. 1. What are the null and alternative hypotheses for the test? A)Hp:P2>0.80and Hy : P <0.80 B) Hy: P=0.80 and H7 : P # 0.80 C) Hy: P<0.80 and H1 : P>0.80 D) Hy: P>0.80 and H1 : P<0.80 2. Whatis the c/:\lecision rule? A) Reject Hy if (P - PQ) / \/P() (1-Po)/n > Zo/2 B) Reject Hy if (P - P0) / Afp (1-p) /7 >z C) Reject Hy if (I/; - Po) / \/P() (1-Pg)/n < “Zn /2 D) Reject Hy if (P - P) / [P0 (1= PQ) /7 <2z, 3. The value of the test statistic is: A) 2.086 B) 1.444 C) -2.086 D) -1.444 4. Which of the following statements is most accurate? A) Fail to reject the null hypothesis at a <0.10 B) Reject the null hypothesis at o = 0.05 C) Reject the null hypothesis at a = 0.10, but not 0.05 D) Reject the null hypothesis at a =0.01 5. If conducting a two-sided test of population means, unknown variance, at level of significance 0.05 based on a sample of size 20, the critical t-value is: A) 1.725 B) 2.093 C) 2.086 D) 1.729
Problem 3: The concentration of a drying compound is critical in gauging the final strength of an adhesive product. The standard deviation of the amount of the compound must be 1,500 parts-per-million (ppm) or less. Based on a sample of 25 mixtures, the standard deviation is 1,700 ppm. Is there reason to believe, at the 10 per cent significance level, that the standard deviation of the drying compound is more than 1,500 ppm? (Note: 1,500 ppm equals a proportion of .0015 and a percent of .000015.) Problem 4: A random sample of 250 business faculty members was asked if there should be a required foreign language course for international business majors. Of these sample members, 170 felt there was a need for a foreign language course. Test the hypothesis that at least 75% of all business faculty members hold this view. Use a= 0.05. Problem 5: A computer information systems manager was in charge of purchasing new battery packs for laptop computers. The choices were narrowed to two available models. Since the two models cost about the same, the manager was interested in determining whether there was a difference in the average time the battery packs would function before needing to be recharged. Based on two independent random samples, the following summary information was computed: Battery Pack Battery Pack Model 1 Model 2 Sample Size 30 30 Sample Mean 6 hours 6.5 hours Population Standard Deviation 1.8 hours 2.6 hours 1. Find a 95% confidence interval for the difference in average functioning time before recharging the two models. 2. Based on the 95% confidence interval, can one conclude that there is a difference in the true average functioning time between the two models of battery packs before recharging? Justify your answer. 3. Find a 90% confidence interval for the difference in average functioning time before recharging the two models. 4. Find a 99% confidence interval for the difference in average functioning time before recharging the two models.
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