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densk [106]
3 years ago
8

For every £1.00 a company spends on advertising, it spends £0.75 on its website. Express the amount spent on advertising to the

amount spent on its website as a ratio in its simplest form.
Mathematics
1 answer:
Mandarinka [93]3 years ago
4 0

Answer:

4:3

Step-by-step explanation:

For every £1.00 a company spends on advertising, it spends £0.75 on its website.

From the above question:

The ratio is given as:

Amount spent on advertising : Amount spent on website

= £1.00 : £0.75

Convert ratio sign to division

= 1/0.75

Multiply Numerator and Denominator by 100

= 1 × 100 / 0.75 × 100

= 100/75

= 4/3

= 4 : 3

Therefore, the ratio in its simplest form = 4:3

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Answer:

a) 52.8-2.02\frac{3.9}{\sqrt{40}}=51.554    

b) 52.8+2.02\frac{3.9}{\sqrt{40}}=54.046    

c) 52.8-2.71\frac{3.9}{\sqrt{40}}=51.129  

d) 52.8+2.71\frac{3.9}{\sqrt{40}}=54.471  

e) Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X =52.8 represent the sample mean for the sample

\mu population mean (variable of interest)

s=3.9 represent the sample standard deviation

n=40 represent the sample size  

a) What is the lower limit of the 95% interval? Give your answer to three decimal places

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=40-1=39

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,39)".And we see that t_{\alpha/2}=2.02

Now we have everything in order to replace into formula (1):

52.8-2.02\frac{3.9}{\sqrt{40}}=51.554    

b) What is the upper limit of the 95% interval? Give your answer to three decimal places

52.8+2.02\frac{3.9}{\sqrt{40}}=54.046    

So on this case the 95% confidence interval would be given by (51.554;54.046)    

c) What is the lower limit of the 99% interval? Give your answer to three decimal places

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,39)".And we see that t_{\alpha/2}=2.71

Now we have everything in order to replace into formula (1):

52.8-2.71\frac{3.9}{\sqrt{40}}=51.129  

d) What is the upper limit of the 99% interval? Give your answer to three decimal places

52.8+2.71\frac{3.9}{\sqrt{40}}=54.471  

So on this case the 99% confidence interval would be given by (51.129;54.471)    

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At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

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