8) In their advertisements, a new diet program would like to claim that their program results in a mean weight loss ($\mu$) of more than $10$ pounds in two weeks. State the null and alternative hypothesis.

9) Suppose we would like to estimate the mean amount of money ($\mu$) spent on books by CS students in a semester. We have the following data from $10$ randomly selected CS students: $X = \$249$ and $S = \$30$. Assume that the amount spent on books by CS students is normally distributed. To compute a 95% conﬁdence for µ, what will we use as the value of our critical point?

10) Installation of a certain hardware takes a random amount of time with a standard deviation of $5$ minutes. A computer technician installs this hardware on $64$ different computers, with the average installation time of $42$ minutes. Compute a $95\%$ conﬁdence interval for the mean installation time.

11) An exit poll of $1000$ randomly selected voters found that $515$ favored candidate A. Is the race too close to call? Answer this question by performing an appropriate test of hypothesis at $1\%$ level of signiﬁcance.

12) Bags of a certain brand of tortilla chips claim to have a net weight of $14$ ounces. The net weights actually vary slightly from bag to bag and are normally distributed with mean $\mu$. A representative of a consumer advocacy group wishes to see if there is any evidence that the mean net weight is less than advertised. For this, the representative randomly selects $16$ bags of this brand and determines the net weight of each. He ﬁnds the sample mean to be $X = 13.82$ and the sample standard deviation to be $S = 0.24$. Use these data to perform an appropriate test of hypothesis at $5\%$ signiﬁcance level.

13) The time needed for college students to complete a certain maze follows a normal distribution with a mean of $45$ seconds. To see if the mean time time $\mu$ (in seconds) is changed by vigorous exercise, we have a group of nine college students exercise vigorously for $30$ minutes and then complete the maze. The sample mean and standard deviation of the collected data is $49.2$ seconds and $3.5$ seconds respectively. Use these data to perform an appropriate test of hypothesis at $5\%$ level of signiﬁcance.

CONFIDENCE INTERVAL

chegendungu

Answered by nhamo manjoro 1 month ago

1)A coin is tossed three times .What is the probability that : a) 1.The first outcome is a head ? 2.The first two outcomes are tails? b) 1.The first two outcomes were heads? 2.The fist two outcomes were tails ? 3.The first outcome was a head and the third was a head ?

Answered by nhamo manjoro 1 month ago

it is assumed that achievement test scores should be correlated with students classroom performance .One would expect that students who consistently perform well in the classroom (tests .quizzes ,etc.)would also perform well on a standardized achievement test (0-100 with 100 indicating high achievement ).A teacher decides to examine this hypothesis .At the end of the academic year ,she computes a correlation between the student's achievement test scores (she purposefully did not look at this data until after she submitted student's grades)and the overall g.p.a for each student computed over the entire year .The data for her class are provided below: G.P.A:3.6 2.7 3.1 4.0 3.2 3.0 3.8 2.6 3.0 2.2 1.7 3.1 2.6 2.9 3.7 achievement:98 96 94 88 91 77 86 71 59 63 84 79 75 72 93

a)Compute the correlation coefficient b)What does this statistic mean concerning the relationship between achievement test performance and g.p.a? c)what percent of the variability is accounted for by the relationship between the two variables and what does this statistic mean ? d)What would be the slope and y -intercept for a regression line based on this data? e)If a student scored a 93 on the achievement test ,what would be their predicted G.P.A?

**SOLUTION**

8)

$$H_0 : \mu = 10$$ $$vs$$ $$H_1 : \mu > 10$$

9)

We want to use a t-interval because the population standard deviation σ is not given. The critical point, in this case, is $t_{n−1,\frac{\alpha}{2}}$. So the critical value is: $$t_{9,0.025} = 2.262$$

10)

Let µ denote the mean installation time (in minutes). We want a $95\%$ conﬁdence interval for $\mu$. The following quantities are given: $1−\alpha = 0.95, n = 64, X = 42$ min, and $\sigma = 5$ min. Since $\sigma$ is known, we will use the $Z$-interval:

$$X ±Z_{\frac{\alpha}{2}}\frac{\sigma}{√n}=42±(1.96)\frac{5}{√64} = 42±1.225 = [40.8,43.2], $$

where $Z_{0.025} = 1.96$ is obtained from the normal table.Thus, the mean installation time is estimated to be between $40.8$ min and $43.2$ min, with $95\%$ conﬁdence.

11)

Let p denote the proportion of voters who favor candidate A. From the data, $\bar{p} = 515/1000 = 0.515$. We would like to test the null hypothesis $H_0 : p = 0.5$ against the two-sided alternative $H_1 : p \neq 0.5$, at $\alpha = 0.01$. Since $n = 1000$ is large, we will do a large-sample $Z$-test. The rejection rejection is $|Z| > Z_{\frac{\alpha}{2}} = 2.576$, using the normal table. The test statistic is:

$$Z =\frac{\bar{p}−p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}=\frac{0.515−0.5}{\sqrt{\frac{0.50.5}{1000}}} = 0.949$$

Since $|Z| = 0.949 < 2.576$, $H_0$ is not rejected. Thus, the race is too close to call.

12)

We need to test the null hypothesis $H_0 : \mu = 14$ against the one-sided alternative hypothesis $H_1 : \mu < 14$, at level $\alpha = 0.05$. Since $n = 16$ is small and the data are normally distributed, we will do a t-test. The rejection region is $T < −t_{n−1,\alpha} = −t_{15,0.05} = −1.753$, using the t-table. The test statistic is:

$$T =\frac{X −\mu_0}{\frac{S}{√n}} =\frac{13.82−14}{\frac{0.24}{\sqrt{16}}}= −3$$

Since $T = −3 < −1.753$, $H_0$ is rejected. Thus, there is evidence at $5\%$ signiﬁcance level that the mean weight is less than advertised.

13) We need to test the null $H_0 : \mu = 45$ against the two-sided alternative $H_1 : \mu \neq45$, at level $\alpha = 0.05$. It is given that $n = 9, X = 49.2$ and $S = 3.5$. Since $n$ is small and the data are normally distributed, we will do a t-test. The rejection region is $|T| > t_{n−1,\frac{\alpha}{2}} = t_{8,0.025} = 2.306$, using the t-table. The test statistic is:

$$T =\frac{X −\mu_0}{\frac{S}{√n}} =\frac{49.2−45}{\frac{3.5}{\sqrt{9}}}= 3.6$$

Since $|T| = 3.6 > 2.306$, $H_0$ is rejected. Thus, there is signiﬁcant evidence at $5\%$ signiﬁcance level that the mean time to complete the maze is changed after the exercise.

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