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Rzqust [24]
3 years ago
8

The US Agency for Healthcare Research and Quality (www.ahrq.gov) reported that in 2010 the mean cost of a stay in a hospital for

American women aged 18-44 was $15,200. A random sample of 400 hospital stays for American women aged 18-44 showed a mean cost of $16,000, with a standard deviation of $5000. Test whether the population mean cost for a hospital stay for American women aged 18-44 has increased since 2010, using the 5% level of significance.
Mathematics
1 answer:
zysi [14]3 years ago
8 0

Answer:

We conclude that the population mean cost for a hospital stay for American women aged 18-44 has increased since 2010.

Step-by-step explanation:

We are given that the US Agency for Healthcare Research and Quality reported that in 2010 the mean cost of a stay in a hospital for American women aged 18-44 was $15,200.

A random sample of 400 hospital stays for American women aged 18-44 showed a mean cost of $16,000, with a standard deviation of $5000.

<u><em>Let </em></u>\mu<u><em> = population mean cost for a hospital stay for American women aged 18-44.</em></u>

SO, <u>Null Hypothesis</u>, H_0 : \mu \leq $15,200   {means that the population mean cost for a hospital stay for American women aged 18-44 has reduced or remains same since 2010}

<u>Alternate Hypothesis</u>, H_A : \mu > $15,200   {means that the population mean cost for a hospital stay for American women aged 18-44 has increased since 2010}

The test statistics that will be used here is <u>One-sample t test</u> <u>statistics</u> as 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 cost of a stay in a hospital for American women aged 18-44 = $16,000

              s = sample standard deviation = $5,000

              n = sample of hospital stays = 400

So, <em><u>test statistics</u></em>  =   \frac{16,000-15,200}{\frac{5,000}{\sqrt{400} } }  ~ t_3_9_9

                               =  3.20

<em>Now at 5% significance level, the t table gives critical value of 1.645 at 399 degree of freedom for right-tailed test. Since our test statistics is more than the critical value of t as 1.645 < 3.20, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the population mean cost for a hospital stay for American women aged 18-44 has increased since 2010.

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