Mathematics, 31.07.2020 16:01 nickname0097

A survey of the average amount of cents off that coupons give was done by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20cents; 70cents; 50cents; 65cents; 30cents; 55cents; 40cents; 40cents; 30cents; 55cents; $1.50; 40cents; 65cents; 40cents. Assume the underlying distribution is approximately normal. Construct a 95% confidence interval for the population mean worth of coupons .
What is the lower bound? ( Round to 3 decimal places )
What is the upper bound? ( Round to 3 decimal places )
What is the error bound? (Round to 3 decimal places)

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Answer from: creeper2737

The lower bound = 35.443

The upper bound = 71.697

The error bound = 18.127

Step-by-step explanation:

We are given that a survey of the average amount of cents off that coupons gives was done by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News.

The following data were collected (X): 20cents; 70cents; 50cents; 65cents; 30cents; 55cents; 40cents; 40cents; 30cents; 55cents; 150 cents; 40cents; 65cents; 40cents.

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

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

where, $\bar X$ = sample mean worth of coupons = $\frac{\sum X}{n}$ = $\frac{750}{14}$ = 53.57 cents

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 31.40 cents

n = sample size = 14

$\mu$ = population mean worth of coupons

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

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

P(-2.16 < t_1_3 < 2.16) = 0.95 {As the critical value of t at 13 degrees of

freedom are -2.16 & 2.16 with P = 2.5%}

P(-2.16 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.16) = 0.95

P( $-2.16 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}{$ < $2.16 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.16 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.16 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.16 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.16 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $53.57-2.16 \times {\frac{31.40}{\sqrt{14} } }$ , $53.57+2.16 \times {\frac{31.40}{\sqrt{14} } }$ ]

= [35.443, 71.697]

Therefore, a 95% confidence interval for the population mean worth of coupons is [35.443, 71.697].

Answer from: Quest

1,407.5 hours

step-by-step explanation:

we first convert all the time to hours and then sum them up.

there are 24 hours in 1 day, in 58 days, there are 24 x 58 = 1,392 hours.

there are 60 minutes in 1 hour, thus, 30 minutes will be equivalent to 30 / 60 = 0.5 hours.

therefore, the number of hours it will take mercury to complete a full rotation is 1,392 + 15 + 0.5 = 1,407.5 hours

Answer from: Quest

$2^{2x+11}=3^{x-31}$

take the logarithm of both sides (the base of the log doesn't matter):

$\ln2^{2x+11}=\ln3^{x-31}$

apply the exponent property - this says that $\ln a^b=b\ln a$ :

$(2x+11)\ln2=(x-31)\ln3$

distribute the log terms on both sides:

$(2\ln2)x+11\ln2=(\ln3)x-31\ln3$

group like terms together:

$(2\ln2-\ln3)x=-11\ln2-31\ln3$

divide through both sides by the coefficient on $x$ :

$x=-\dfrac{11\ln2+31\ln3}{2\ln2-\ln3}$

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like:

$x=-\dfrac{\ln2^{11}+\ln3^{31}}{\ln2^2-\ln3}$

$x=-\dfrac{\ln2^{11}3^{31}}{\ln\frac43}$

$x=-\log_{4/3}2^{11}3^{31}$

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