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4 years ago

Emily thinks that -11/6 - (-2/3) is -5/2. Identify the error ghat emily made. Then correct Emily’s error and find the correct va

lue.

Mathematics

1 answer:

Amanda [17]4 years ago

7 0

$\bf \stackrel{\textit{she subtracted }\frac{2}{3}\textit{ instead of adding it}}{-\cfrac{11}{6}-\left(\cfrac{2}{3} \right)\impliedby \stackrel{LCD}{6}\implies \cfrac{-(1)11~~-~~(2)2}{6}}\implies \cfrac{-11~~-~~4}{6}
\\\\\\
\cfrac{-15}{6}\implies \stackrel{simplified}{-\cfrac{5}{2}}\impliedby \textit{that's correct, assuming is }\frac{2}{3}~~but~is~~-\frac{2}{3}\\\\
-------------------------------$

$\bf \stackrel{\textit{recall that minus times minus is plus}}{-\cfrac{11}{6}-\left(-\cfrac{2}{3} \right)\implies -\cfrac{11}{6}+\cfrac{2}{3}}\implies \cfrac{-(1)11+(2)2}{6}
\\\\\\
\cfrac{-11+4}{6}\implies -\cfrac{7}{6}\quad \checkmark$

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The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

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Complete Question

The complete question is shown on the first uploaded image

Answer:

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis