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Anuta_ua [19.1K]
3 years ago
7

PLS help and explain these problems

Mathematics
1 answer:
vichka [17]3 years ago
6 0
For 19 I got z=15. And for 20 I got k=8.6
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A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.
nika2105 [10]

Answer:

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

\sigma=12 represent the population deviation

n=76 represent the sample size  

Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

Replacing we got:

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0
3 years ago
Please help, due tonight
ryzh [129]

Answer:

There's an app call ( m a t h w a y) that would give u the answer in sec

8 0
2 years ago
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

<em>Let </em>\mu<em> = true average percentage of organic matter in such soil</em>

SO, <u>Null Hypothesis</u>, H_0 : \mu = 3%   {means that the true average percentage of organic matter in such soil is equal to 3%}

<u>Alternate Hypothesis</u>, H_A : \mu \neq 3%   {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is <u>One-sample t test statistics</u> because we don't know about the population standard deviation;

                           T.S.  = \frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }  ~ t_n_-_1

where,  \bar X = sample mean amount of organic matter = 2.481%

              s = sample standard deviation = 1.616%

              n = sample of soil specimens = 30

So, <u><em>test statistics</em></u>  =  \frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }  ~ t_2_9

                               =  -1.759

<u></u>

<u>Now, P-value of the test statistics is given by;</u>

       P-value = P( t_2_9 > -1.759) = <u>0.046</u> or 4.6%

  • If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.
  • If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

<em>Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0
3 years ago
Which table of values corresponds to the graph below?
Karolina [17]

The correct answer is the second answer choice.  It is the one that starts with (0,-2).

6 0
3 years ago
Read 2 more answers
What's 5+5×5+5 equal
gogolik [260]
5+5=10
5x5=25
 plus 5=35
8 0
2 years ago
Read 2 more answers
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