Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test of mu = 100 vs not = 100 Variable N Mean StDev SE Mean 95% CI (Lower) 95% CI (Upper) T X 17 99.13 4.60 i (a) How many degrees of freedom are there on the t-statistic? 1 (b) What is your conclusion if a = 0.05? (c) What is your conclusion if the hypothesis is Ho:μ = 100 versus H₁: > 100?

Assume that you want to test the claim that the paired smaple data come from a population for which the mean difference is population mean d=0. Compute the value of the t test statistic.  Round intermediate calculations to four decimal places as need and final answers to three decimal places as needed x 9 6 7 5 12 y 6 8 3 6 7

A runner states that his average time for running a mile is 6.5 minutes.  Given the following times from 8 recent mile runs, test the claim that his  actual average is not what he states it to be at the a = 0.05 level of  significance.  6.25    6.6    6.8    7.2    6.3    7.0    6.5    6.75

mg mg Calcium levels in people are normally distributed with a mean of 9.5 and a standard deviation of 0.3 dL dL Individuals with calcium levels in the bottom 15% of the population are considered to have low calcium levels. Find the calcium level that is the borderline between low calcium levels and those not considered low. Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place. mg dL X

Q.4 F

hand written asap

Average speed on a freeway is  80  mph.   Among all vehicles 65% drive 8 miles above the speed limit of 70 mph.   34 What fraction drive within          ± 5 mph from the mean speed? a 0.7327             b 0.6978             c 0.6646             d 0.6330

The authors of the paper "Evaluating Existing Movement Hypotheses in Linear Systems Using Larval Stream Salamanders"+ investigated whether water temperature was related to how far a salamander would swim and whether it would swim upstream or downstream. Data for 14 streams with different mean water temperatures where salamander larvae were released are given (approximated from a graph that appeared in the paper). The two variables of interest are x = mean water temperature (°C) and y = net directionality, which was defined as the difference in the relative frequency of the released salamander larvae moving upstream and the relative frequency of released salamander larvae moving downstream. A positive value of net directionality means a higher proportion were moving upstream than downstream. A negative value of net directionality means a higher proportion were moving downstream than upstream. Mean Temperature (x) Net Directionality (y) 6.17 -0.08 8.06 0.25 8.62 -0.14 10.56 0.00 12.45…

Use the output below to determine if there is a significant difference in the hour spent emailing between the two group if T0 =0.290

Given the below output, interpret R-squared. Note: the response variable is Charges Coefficients: (Intercept) age bmi regionnorthwest -979.86 regionsoutheast 64.83 regionsouthwest -1562.19 Estimate Std. Error t value Pr(>|t|) -5510.16 1814.88 -3.036 0.00244 ** 22.30 10.893 < 2e-16 *** 53.62 5.994 2.64e-09 *** 893.36 -1.097 0.27291 897.84 0.072 0.94245 896.58 -1.742 0.08167 . 242.97 321.37 signif. codes: 0 ***' 0.001 **' 0.01 ** 0.05 .' 0.1'1 Residual standard error: 11380 on 1332 degrees of freedom Multiple R-squared: 0.1203, F-statistic: 36.44 on 5 and 1332 DF, p-value: < 2.2e-16 Adjusted R-squared: 0.117 O a) 11.7% of the variation in Charges can be explained by the regression model b) 12.03% of the variation in Charges can be explained by the regression model c) 12.03% of the variation in Charges can be explained by the variation of the regression model d) 11.7% of the variation in Charges can be explained by the variation of the regression model