What is 12 over 3 as a proper fraction?

In 1998, as an advertising campaign, the Nabisco Company announced a "1000 Chips Challenge," claiming that every 18-ounce bag of

Phoenix [80]

Answer:

A 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

Step-by-step explanation:

We are given that statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies and counted the chocolate chips.

Some of their data are given below; 1219, 1214, 1087, 1200, 1419, 1121, 1325, 1345, 1244, 1258, 1356, 1132, 1191, 1270, 1295, 1135.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of chocolate chips = $\frac{\sum X}{n}$ = 1238.2

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 94.3

n = sample of car drivers = 16

$\mu$ = population mean number of chips in a bag

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $1238.2-2.131 \times {\frac{94.3}{\sqrt{16} } }$ , $1238.2+2.131 \times {\frac{94.3}{\sqrt{16} } }$ ]

= [1187.96, 1288.44]

Therefore, a 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

7 0

3 years ago

Trevor thinks that left-handed pitchers in a baseball league are more likely to strike out batters than right-handed pitchers ar

Scilla [17]

Answer:

C. H0: PL - PR = 0

Ha: PL - PR > 0

The null hypothesis for this case is that left-handed pitchers in a baseball league are not more likely to strike out batters than right-handed pitchers are. That is the proportion of at-bats resulting in a strikeout is significantly equal for both left-handed and right-handed pitchers.

H0: PL - PR = 0

The alternative hypothesis is that the proportion of at-bats resulting in a strikeout is significantly higher for left-handed pitchers

Ha: PL - PR > 0

PL - proportion of at-bats resulting in a strikeout for left-handed pitchers

PR - proportion of at-bats resulting in a strikeout for right-handed pitchers

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

The null hypothesis for this case is that left-handed pitchers in a baseball league are not more likely to strike out batters than right-handed pitchers are. That is the proportion of at-bats resulting in a strikeout is significantly equal for both left-handed and right-handed pitchers.

H0: PL - PR = 0

The alternative hypothesis is that the proportion of at-bats resulting in a strikeout is significantly higher for left-handed pitchers

Ha: PL - PR > 0

PL - proportion of at-bats resulting in a strikeout for left-handed pitchers

PR - proportion of at-bats resulting in a strikeout for right-handed pitchers

8 0

3 years ago

What is the answer to -3x+2=-7

ycow [4]

Answer:

x=3

Step-by-step explanation:

-3x=-7-2=-9

x=-9/-3=3

4 0

3 years ago

Read 2 more answers

Using the Taylor rule, if inflation is 1 percent, desired inflation is 2 percent, and output is 2 percentage points below potent

antoniya [11.8K]

Answer:

1.5%

Step-by-step explanation:

According to the Taylor rule, the federal funds rate targeted by the Fed is:

$FFR = 2+AI+0.5*(AI-DI)+0.5PD$

Where AI is the actual inflation (1%), DI is the desired inflation (2%) and PD is the deviation from potential (2%):

$FFR = 2+1+0.5*(1-2)+0.5*(-2)\\FFR=1.5\%$

The Fed should target a federal funds rate of 1.5%.

6 0

3 years ago

Using traditional methods, it takes 10.1 hours to receive a basic driving license. A new license training method using Computer

Law Incorporation [45]

Answer:

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis: $\mu =10.1$

Alternative hypothesis: $\mu \neq 10.1$

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$

$p_v =2*P(t_{(11)}>0.990)=0.343$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=10.5$ represent the sample mean

$s=1.4$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =10.1$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis: $\mu =10.1$

Alternative hypothesis: $\mu \neq 10.1$

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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:

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=12-1=11$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(11)}>0.990)=0.343$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.

7 0

3 years ago