Calculate a 95% confidence interval for the effect of a one unit increase in exercise on resting heart rate for a non-smoker (Smoke=0). That is, creat a 95% confidence interval for B2. Use 1.96 as the...


Calculate a 95% confidence interval for the effect of a one unit increase in exercise on resting<br>heart rate for a non-smoker (Smoke=0). That is, creat a 95% confidence interval for B2. Use 1.96<br>as the 95% multiplier.<br>What is the lower limit of this 95% confidence interval?<br>

Extracted text: Calculate a 95% confidence interval for the effect of a one unit increase in exercise on resting heart rate for a non-smoker (Smoke=0). That is, creat a 95% confidence interval for B2. Use 1.96 as the 95% multiplier. What is the lower limit of this 95% confidence interval?
An observational study was conducted where subjects were randomly sampled and then had<br>their resting heart rate recorded, as well as their smoking status (0 for non-smoker and 1 for<br>smoker) and how much they exercise on average each day (in hours).<br>A linear regression model is fit where we have response variable of resting heart rate and<br>explanatory variables of smoking status (0 for non-smoker and 1 for smoker) and exercise<br>amount per day in hours, along with an interaction between smoking status and exercise<br>amount.<br>The output is below:<br>Coefficients:<br>Estimate Std. Error t value Pr(>t)<br>(Intercept)<br>84.8172<br>1.9553<br>43.377<br><2e-16 ***<br>Smoke<br>-2.2645<br>5.0665<br>-0.447<br>0.6551<br>Exercise<br>-7.3684<br>0.8075<br>-9.125<br><2e-16 ***<br>Smoke: Exercise<br>1.9562<br>2.5510<br>0.767<br>0.4442<br>Signif. codes:<br>O **** 0.001 (*** 0.01 *' 0.05 .' 0.1 '<br>1<br>Residual standard error: 8.396 on 228 degrees of freedom<br>Adjusted R-squared:<br>Multiple R-squared: 0.2971,<br>F-statistic: 32.13 on 3 and 228 DF, p-value: < 2.2e-16<br>0.2879<br>The model that was fit is:<br>Rest; = Bo + B, Smoke; + B2Exercise; + B3 Smoke; * Exercise; + e;<br>

Extracted text: An observational study was conducted where subjects were randomly sampled and then had their resting heart rate recorded, as well as their smoking status (0 for non-smoker and 1 for smoker) and how much they exercise on average each day (in hours). A linear regression model is fit where we have response variable of resting heart rate and explanatory variables of smoking status (0 for non-smoker and 1 for smoker) and exercise amount per day in hours, along with an interaction between smoking status and exercise amount. The output is below: Coefficients: Estimate Std. Error t value Pr(>t) (Intercept) 84.8172 1.9553 43.377 <2e-16 ***="" smoke="" -2.2645="" 5.0665="" -0.447="" 0.6551="" exercise="" -7.3684="" 0.8075="" -9.125=""><2e-16 ***="" smoke:="" exercise="" 1.9562="" 2.5510="" 0.767="" 0.4442="" signif.="" codes:="" o="" ****="" 0.001="" (***="" 0.01="" *'="" 0.05="" .'="" 0.1="" '="" 1="" residual="" standard="" error:="" 8.396="" on="" 228="" degrees="" of="" freedom="" adjusted="" r-squared:="" multiple="" r-squared:="" 0.2971,="" f-statistic:="" 32.13="" on="" 3="" and="" 228="" df,="" p-value:="">< 2.2e-16="" 0.2879="" the="" model="" that="" was="" fit="" is:="" rest;="Bo" +="" b,="" smoke;="" +="" b2exercise;="" +="" b3="" smoke;="" *="" exercise;="" +="">

Jun 02, 2022
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