Mathematics, 03.01.2020 04:31 ooEVAoo

All fifth-grade students are given a test on academic achievement in new york state. suppose the mean score is 70 for the entire state. a random sample of fifth-grade students is selected from long island. below are the scores in this sample from a normal population. 82 94 66 87 68 85 68 84 70 83 65 70 83 71 82 72 73 81 76 74 a. construct a 95% confidence interval for the population mean score on long island. b. construct a 90% confidence interval for the population standard deviation of the scores on long island. c. a teacher at a long island high school claims that the mean score on long island is higher than the mean for new york state. conduct a test to see if this claim is reasonable using α = 0.01. d. find the p-value of the test.

Answers

a) The 90% confidence interval would be given by (73.56;79.84)

b) The 90% confidence interval for the deviation would be $6.44 \leq \sigma \leq 11.116$ .

c) If we compare the p value and the significance level given $\alpha=0.01$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance. So the claim of the teacher makes sense.

d) Since is a one-side upper test the p value would given by:

$p_v =P(t_{19}3.69)=0.00078$

Step-by-step explanation:

Data given: 82 94 66 87 68 85 68 84 70 83 65 70 83 71 82 72 73 81 76 74

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

And we got the following results:

$\bar X=76.7$ represent the sample mean

$\mu$ population mean (variable of interest)

s=8.112 represent the population standard deviation

n=20 represent the sample size

90% confidence interval

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are given by:

df=n-1=20-1=19

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,19)".And we see that $t_{\alpha/2}=1.73$

Now we have everything in order to replace into formula (1):

$76.7-1.73\frac{8.112}{\sqrt{20}}=73.56$

$76.7+1.73\frac{8.112}{\sqrt{20}}=79.84$

So on this case the 90% confidence interval would be given by (73.56;79.84)

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=20-1=19

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,19)" "=CHISQ.INV(0.95,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=30.143$

$\chi^2_{1- \alpha/2}=10.117$

And replacing into the formula for the interval we got:

$\frac{(19)(8.112^2)}{30.143} \leq \sigma^2 \leq \frac{(19)(8.112^2)}{10.117}$

$41.472 \leq \sigma^2 \leq 123.574$

So the 90% confidence interval for the deviation would be $6.44 \leq \sigma \leq 11.116$ .

Part c

Null hypothesis: $\mu \leq 70$

Alternative hypothesis: $\mu 70$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{76.7-70}{\frac{8.112}{\sqrt{20}}}=3.69$

Part d

P-value

Since is a one-side upper test the p value would given by:

$p_v =P(t_{19}3.69)=0.00078$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance.

Answer from: QueenNerdy889

(a) 95% confidence interval for the population mean score on Long Island is (73, 80.40)

(b) 90% confidence interval for the population standard deviation of the scores on Long Island is (4.85, 10.97)

(d) p-value is 0.01

Step-by-step explanation:

From the data values from Long Island,

Mean is 76.7 and standard deviation is 7.91

Confidence Interval (CI) = mean + or - (t × sd)/√n

(a) mean = 76.7, sd = 7.91, n = 20, degree of freedom = n-1 = 20-1 = 19, t-value corresponding to 19 degrees of freedom and 95% confidence level is 2.093

Lower bound = 76.7 - (2.093×7.91)/√20 = 76.7 - 3.70 = 73

Upper bound = 76.7 + (2.093×7.91)/√20 = 76.7 + 3.70 = 80.40

95% CI is (73, 80.40)

(b) CI = sd + or - (t×sd)/√n

sd = 7.91, t-value corresponding to 19 degrees of freedom and 90% confidence level is 1.729

Lower bound = 7.91 - (1.729×7.91)/√20 = 7.91 - 3.06 = 4.85

Upper bound = 7.91 + (1.729×7.91)/√20 = 7.91 + 3.06 = 10.97

90% CI is (4.85, 10.97)

Alternate hypothesis: The mean score on Long Island is greater than 70

Z = (sample mean - population mean)/(sd/√n) = (70 - 76.7)/(7.91/√20) = -6.7/1.77 = -3.79

Using 0.01 significance level, the critical value is 2.326

Since -3.79 is less than 2.326, reject the null hypothesis. The teacher's claim is reasonable